3.435 \(\int \frac {x^{3/2} (a+b x^2)^2}{(c+d x^2)^3} \, dx\)

Optimal. Leaf size=402 \[ \frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}+\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\sqrt {x} \left (\frac {3 a^2 d}{c}+10 a b-\frac {45 b^2 c}{d}\right )}{16 c d^2}-\frac {x^{5/2} (b c-a d) (3 a d+13 b c)}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {x^{5/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[Out]

1/4*(-a*d+b*c)^2*x^(5/2)/c/d^2/(d*x^2+c)^2-1/16*(-a*d+b*c)*(3*a*d+13*b*c)*x^(5/2)/c^2/d^2/(d*x^2+c)+1/64*(-3*a
^2*d^2-10*a*b*c*d+45*b^2*c^2)*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(7/4)/d^(13/4)*2^(1/2)-1/64*(-3*a^2*
d^2-10*a*b*c*d+45*b^2*c^2)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(7/4)/d^(13/4)*2^(1/2)+1/128*(-3*a^2*d^
2-10*a*b*c*d+45*b^2*c^2)*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(7/4)/d^(13/4)*2^(1/2)-1/128*
(-3*a^2*d^2-10*a*b*c*d+45*b^2*c^2)*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))/c^(7/4)/d^(13/4)*2^(1
/2)-1/16*(10*a*b-45*b^2*c/d+3*a^2*d/c)*x^(1/2)/c/d^2

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Rubi [A]  time = 0.33, antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {463, 457, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}+\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\sqrt {x} \left (\frac {3 a^2 d}{c}+10 a b-\frac {45 b^2 c}{d}\right )}{16 c d^2}-\frac {x^{5/2} (b c-a d) (3 a d+13 b c)}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {x^{5/2} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(3/2)*(a + b*x^2)^2)/(c + d*x^2)^3,x]

[Out]

-((10*a*b - (45*b^2*c)/d + (3*a^2*d)/c)*Sqrt[x])/(16*c*d^2) + ((b*c - a*d)^2*x^(5/2))/(4*c*d^2*(c + d*x^2)^2)
- ((b*c - a*d)*(13*b*c + 3*a*d)*x^(5/2))/(16*c^2*d^2*(c + d*x^2)) + ((45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*Arc
Tan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(7/4)*d^(13/4)) - ((45*b^2*c^2 - 10*a*b*c*d - 3*a^2*
d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(7/4)*d^(13/4)) + ((45*b^2*c^2 - 10*a*b*c*d
- 3*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(7/4)*d^(13/4)) - ((45*
b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(7
/4)*d^(13/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^{3/2} \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx &=\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int \frac {x^{3/2} \left (\frac {1}{2} \left (-8 a^2 d^2+5 (b c-a d)^2\right )-4 b^2 c d x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d^2}\\ &=\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \int \frac {x^{3/2}}{c+d x^2} \, dx}{32 c^2 d^2}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{32 c d^3}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{16 c d^3}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{32 c^{3/2} d^3}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx,x,\sqrt {x}\right )}{32 c^{3/2} d^3}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^{3/2} d^{7/2}}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^{3/2} d^{7/2}}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}+2 x}{-\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}-2 x}{-\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}\\ &=\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \sqrt {x}}{16 c^2 d^3}+\frac {(b c-a d)^2 x^{5/2}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (13 b c+3 a d) x^{5/2}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{7/4} d^{13/4}}+\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}-\frac {\left (45 b^2 c^2-10 a b c d-3 a^2 d^2\right ) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {d} x\right )}{64 \sqrt {2} c^{7/4} d^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 361, normalized size = 0.90 \[ \frac {\frac {8 \sqrt [4]{d} \sqrt {x} \left (a^2 d^2-18 a b c d+17 b^2 c^2\right )}{c \left (c+d x^2\right )}+\frac {\sqrt {2} \left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{c^{7/4}}-\frac {\sqrt {2} \left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{c^{7/4}}+\frac {2 \sqrt {2} \left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{c^{7/4}}-\frac {2 \sqrt {2} \left (-3 a^2 d^2-10 a b c d+45 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{c^{7/4}}-\frac {32 \sqrt [4]{d} \sqrt {x} (b c-a d)^2}{\left (c+d x^2\right )^2}+256 b^2 \sqrt [4]{d} \sqrt {x}}{128 d^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(3/2)*(a + b*x^2)^2)/(c + d*x^2)^3,x]

[Out]

(256*b^2*d^(1/4)*Sqrt[x] - (32*d^(1/4)*(b*c - a*d)^2*Sqrt[x])/(c + d*x^2)^2 + (8*d^(1/4)*(17*b^2*c^2 - 18*a*b*
c*d + a^2*d^2)*Sqrt[x])/(c*(c + d*x^2)) + (2*Sqrt[2]*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*ArcTan[1 - (Sqrt[2]
*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(7/4) - (2*Sqrt[2]*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(
1/4)*Sqrt[x])/c^(1/4)])/c^(7/4) + (Sqrt[2]*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)
*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(7/4) - (Sqrt[2]*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*
c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(7/4))/(128*d^(13/4))

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fricas [B]  time = 0.60, size = 1420, normalized size = 3.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

1/64*(4*(c*d^5*x^4 + 2*c^2*d^4*x^2 + c^3*d^3)*(-(4100625*b^8*c^8 - 3645000*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^
2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*
c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(1/4)*arctan((sqrt(c^4*d^6*sqrt(-(4100625*b^8*c^8 - 3645000*a*b^7*c^7*d + 1215
00*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^5 + 540*a^6*b^2*c^2*
d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13)) + (2025*b^4*c^4 - 900*a*b^3*c^3*d - 170*a^2*b^2*c^2*d^2 + 60*
a^3*b*c*d^3 + 9*a^4*d^4)*x)*c^5*d^10*(-(4100625*b^8*c^8 - 3645000*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 54900
0*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 8
1*a^8*d^8)/(c^7*d^13))^(3/4) + (45*b^2*c^7*d^10 - 10*a*b*c^6*d^11 - 3*a^2*c^5*d^12)*sqrt(x)*(-(4100625*b^8*c^8
- 3645000*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b
^3*c^3*d^5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(3/4))/(4100625*b^8*c^8 - 364500
0*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^
5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)) + (c*d^5*x^4 + 2*c^2*d^4*x^2 + c^3*d^3)*(-(4100625*b
^8*c^8 - 3645000*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600
*a^5*b^3*c^3*d^5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(1/4)*log(c^2*d^3*(-(41006
25*b^8*c^8 - 3645000*a*b^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 3
6600*a^5*b^3*c^3*d^5 + 540*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(1/4) - (45*b^2*c^2 -
10*a*b*c*d - 3*a^2*d^2)*sqrt(x)) - (c*d^5*x^4 + 2*c^2*d^4*x^2 + c^3*d^3)*(-(4100625*b^8*c^8 - 3645000*a*b^7*c^
7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^5 + 540*a^
6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(1/4)*log(-c^2*d^3*(-(4100625*b^8*c^8 - 3645000*a*b
^7*c^7*d + 121500*a^2*b^6*c^6*d^2 + 549000*a^3*b^5*c^5*d^3 - 42650*a^4*b^4*c^4*d^4 - 36600*a^5*b^3*c^3*d^5 + 5
40*a^6*b^2*c^2*d^6 + 1080*a^7*b*c*d^7 + 81*a^8*d^8)/(c^7*d^13))^(1/4) - (45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*
sqrt(x)) + 4*(32*b^2*c*d^2*x^4 + 45*b^2*c^3 - 10*a*b*c^2*d - 3*a^2*c*d^2 + (81*b^2*c^2*d - 18*a*b*c*d^2 + a^2*
d^3)*x^2)*sqrt(x))/(c*d^5*x^4 + 2*c^2*d^4*x^2 + c^3*d^3)

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giac [A]  time = 0.43, size = 426, normalized size = 1.06 \[ \frac {2 \, b^{2} \sqrt {x}}{d^{3}} - \frac {\sqrt {2} {\left (45 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{2} d^{4}} - \frac {\sqrt {2} {\left (45 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{2} d^{4}} - \frac {\sqrt {2} {\left (45 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{2} d^{4}} + \frac {\sqrt {2} {\left (45 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} - 10 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{2} d^{4}} + \frac {17 \, b^{2} c^{2} d x^{\frac {5}{2}} - 18 \, a b c d^{2} x^{\frac {5}{2}} + a^{2} d^{3} x^{\frac {5}{2}} + 13 \, b^{2} c^{3} \sqrt {x} - 10 \, a b c^{2} d \sqrt {x} - 3 \, a^{2} c d^{2} \sqrt {x}}{16 \, {\left (d x^{2} + c\right )}^{2} c d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")

[Out]

2*b^2*sqrt(x)/d^3 - 1/64*sqrt(2)*(45*(c*d^3)^(1/4)*b^2*c^2 - 10*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^
2)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt(x))/(c/d)^(1/4))/(c^2*d^4) - 1/64*sqrt(2)*(45*(c*d^3)^(1/4
)*b^2*c^2 - 10*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*s
qrt(x))/(c/d)^(1/4))/(c^2*d^4) - 1/128*sqrt(2)*(45*(c*d^3)^(1/4)*b^2*c^2 - 10*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3
)^(1/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^2*d^4) + 1/128*sqrt(2)*(45*(c*d^3)^(1/4)*
b^2*c^2 - 10*(c*d^3)^(1/4)*a*b*c*d - 3*(c*d^3)^(1/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d)
)/(c^2*d^4) + 1/16*(17*b^2*c^2*d*x^(5/2) - 18*a*b*c*d^2*x^(5/2) + a^2*d^3*x^(5/2) + 13*b^2*c^3*sqrt(x) - 10*a*
b*c^2*d*sqrt(x) - 3*a^2*c*d^2*sqrt(x))/((d*x^2 + c)^2*c*d^3)

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maple [A]  time = 0.02, size = 568, normalized size = 1.41 \[ \frac {a^{2} x^{\frac {5}{2}}}{16 \left (d \,x^{2}+c \right )^{2} c}-\frac {9 a b \,x^{\frac {5}{2}}}{8 \left (d \,x^{2}+c \right )^{2} d}+\frac {17 b^{2} c \,x^{\frac {5}{2}}}{16 \left (d \,x^{2}+c \right )^{2} d^{2}}-\frac {3 a^{2} \sqrt {x}}{16 \left (d \,x^{2}+c \right )^{2} d}-\frac {5 a b c \sqrt {x}}{8 \left (d \,x^{2}+c \right )^{2} d^{2}}+\frac {13 b^{2} c^{2} \sqrt {x}}{16 \left (d \,x^{2}+c \right )^{2} d^{3}}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{64 c^{2} d}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{64 c^{2} d}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a^{2} \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{128 c^{2} d}+\frac {5 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{32 c \,d^{2}}+\frac {5 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{32 c \,d^{2}}+\frac {5 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a b \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{64 c \,d^{2}}-\frac {45 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{64 d^{3}}-\frac {45 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{64 d^{3}}-\frac {45 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b^{2} \ln \left (\frac {x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{128 d^{3}}+\frac {2 b^{2} \sqrt {x}}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x^2+a)^2/(d*x^2+c)^3,x)

[Out]

2*b^2/d^3*x^(1/2)+1/16/(d*x^2+c)^2/c*x^(5/2)*a^2-9/8/d/(d*x^2+c)^2*x^(5/2)*a*b+17/16/d^2/(d*x^2+c)^2*c*x^(5/2)
*b^2-3/16/d/(d*x^2+c)^2*x^(1/2)*a^2-5/8/d^2/(d*x^2+c)^2*x^(1/2)*a*b*c+13/16/d^3/(d*x^2+c)^2*x^(1/2)*b^2*c^2+3/
64/d/c^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+5/32/d^2/c*(c/d)^(1/4)*2^(1/2)*arctan(2
^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b-45/64/d^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2+3/12
8/d/c^2*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d
)^(1/2)))*a^2+5/64/d^2/c*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*2^(
1/2)*x^(1/2)+(c/d)^(1/2)))*a*b-45/128/d^3*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2))/(
x-(c/d)^(1/4)*2^(1/2)*x^(1/2)+(c/d)^(1/2)))*b^2+3/64/d/c^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1
/2)+1)*a^2+5/32/d^2/c*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-45/64/d^3*(c/d)^(1/4)*2^(1
/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2

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maxima [A]  time = 2.44, size = 374, normalized size = 0.93 \[ \frac {{\left (17 \, b^{2} c^{2} d - 18 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {5}{2}} + {\left (13 \, b^{2} c^{3} - 10 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} \sqrt {x}}{16 \, {\left (c d^{5} x^{4} + 2 \, c^{2} d^{4} x^{2} + c^{3} d^{3}\right )}} + \frac {2 \, b^{2} \sqrt {x}}{d^{3}} - \frac {\frac {2 \, \sqrt {2} {\left (45 \, b^{2} c^{2} - 10 \, a b c d - 3 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (45 \, b^{2} c^{2} - 10 \, a b c d - 3 \, a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (45 \, b^{2} c^{2} - 10 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (45 \, b^{2} c^{2} - 10 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{128 \, c d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

1/16*((17*b^2*c^2*d - 18*a*b*c*d^2 + a^2*d^3)*x^(5/2) + (13*b^2*c^3 - 10*a*b*c^2*d - 3*a^2*c*d^2)*sqrt(x))/(c*
d^5*x^4 + 2*c^2*d^4*x^2 + c^3*d^3) + 2*b^2*sqrt(x)/d^3 - 1/128*(2*sqrt(2)*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2
)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c
)*sqrt(d))) + 2*sqrt(2)*(45*b^2*c^2 - 10*a*b*c*d - 3*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2
*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(45*b^2*c^2 - 10*a*b*c*d -
3*a^2*d^2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(3/4)*d^(1/4)) - sqrt(2)*(45*b^2*c^2
- 10*a*b*c*d - 3*a^2*d^2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(3/4)*d^(1/4)))/(c*d^
3)

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mupad [B]  time = 0.44, size = 1236, normalized size = 3.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(a + b*x^2)^2)/(c + d*x^2)^3,x)

[Out]

(2*b^2*x^(1/2))/d^3 - (x^(1/2)*((3*a^2*d^2)/16 - (13*b^2*c^2)/16 + (5*a*b*c*d)/8) - (x^(5/2)*(a^2*d^3 + 17*b^2
*c^2*d - 18*a*b*c*d^2))/(16*c))/(c^2*d^3 + d^5*x^4 + 2*c*d^4*x^2) + (atan(((((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*
c*d)^2/(64*(-c)^(7/4)*d^(13/4)) - (x^(1/2)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d +
60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)*1i)/(64*(-c)^(7/4)*d^(13/4)) - (((3*a^2*
d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2/(64*(-c)^(7/4)*d^(13/4)) + (x^(1/2)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c
^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)*1i)/(64*(-c)^(
7/4)*d^(13/4)))/((((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2/(64*(-c)^(7/4)*d^(13/4)) - (x^(1/2)*(9*a^4*d^4 + 20
25*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 +
10*a*b*c*d))/(64*(-c)^(7/4)*d^(13/4)) + (((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2/(64*(-c)^(7/4)*d^(13/4)) + (
x^(1/2)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*
a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d))/(64*(-c)^(7/4)*d^(13/4))))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)*1i)/(32*(
-c)^(7/4)*d^(13/4)) + (atan((((((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2*1i)/(64*(-c)^(7/4)*d^(13/4)) - (x^(1/2
)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^
2 - 45*b^2*c^2 + 10*a*b*c*d))/(64*(-c)^(7/4)*d^(13/4)) - ((((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2*1i)/(64*(-
c)^(7/4)*d^(13/4)) + (x^(1/2)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d
^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d))/(64*(-c)^(7/4)*d^(13/4)))/(((((3*a^2*d^2 - 45*b^2*c^
2 + 10*a*b*c*d)^2*1i)/(64*(-c)^(7/4)*d^(13/4)) - (x^(1/2)*(9*a^4*d^4 + 2025*b^4*c^4 - 170*a^2*b^2*c^2*d^2 - 90
0*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)*1i)/(64*(-c)^(7/4)*d^(13/
4)) + ((((3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d)^2*1i)/(64*(-c)^(7/4)*d^(13/4)) + (x^(1/2)*(9*a^4*d^4 + 2025*b^4
*c^4 - 170*a^2*b^2*c^2*d^2 - 900*a*b^3*c^3*d + 60*a^3*b*c*d^3))/(64*c^2*d^3))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b
*c*d)*1i)/(64*(-c)^(7/4)*d^(13/4))))*(3*a^2*d^2 - 45*b^2*c^2 + 10*a*b*c*d))/(32*(-c)^(7/4)*d^(13/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x**2+a)**2/(d*x**2+c)**3,x)

[Out]

Timed out

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