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

Optimal. Leaf size=364 \[ -\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{11/4} d^{9/4}}+\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{11/4} d^{9/4}}-\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{11/4} d^{9/4}}+\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt {2} c^{11/4} d^{9/4}}-\frac {\sqrt {x} (7 a d+9 b c) (b c-a d)}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\sqrt {x} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[Out]

-1/64*(21*a^2*d^2+6*a*b*c*d+5*b^2*c^2)*arctan(1-d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(11/4)/d^(9/4)*2^(1/2)+1/64
*(21*a^2*d^2+6*a*b*c*d+5*b^2*c^2)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1/4))/c^(11/4)/d^(9/4)*2^(1/2)-1/128*(21
*a^2*d^2+6*a*b*c*d+5*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^(11/4)/d^(9/4)*2^(1/2)+1
/128*(21*a^2*d^2+6*a*b*c*d+5*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^(11/4)/d^(9/4)*2
^(1/2)+1/4*(-a*d+b*c)^2*x^(1/2)/c/d^2/(d*x^2+c)^2-1/16*(-a*d+b*c)*(7*a*d+9*b*c)*x^(1/2)/c^2/d^2/(d*x^2+c)

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Rubi [A]  time = 0.30, antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {463, 457, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{11/4} d^{9/4}}+\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{64 \sqrt {2} c^{11/4} d^{9/4}}-\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{32 \sqrt {2} c^{11/4} d^{9/4}}+\frac {\left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt {2} c^{11/4} d^{9/4}}-\frac {\sqrt {x} (7 a d+9 b c) (b c-a d)}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\sqrt {x} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[x])/(4*c*d^2*(c + d*x^2)^2) - ((b*c - a*d)*(9*b*c + 7*a*d)*Sqrt[x])/(16*c^2*d^2*(c + d*x^2
)) - ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(11/4)
*d^(9/4)) + ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c
^(11/4)*d^(9/4)) - ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[
d]*x])/(64*Sqrt[2]*c^(11/4)*d^(9/4)) + ((5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(
1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(11/4)*d^(9/4))

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 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 {\left (a+b x^2\right )^2}{\sqrt {x} \left (c+d x^2\right )^3} \, dx &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {\int \frac {\frac {1}{2} \left (-8 a^2 d^2+(b c-a d)^2\right )-4 b^2 c d x^2}{\sqrt {x} \left (c+d x^2\right )^2} \, dx}{4 c d^2}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \int \frac {1}{\sqrt {x} \left (c+d x^2\right )} \, dx}{32 c^2 d^2}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (5 b^2 c^2+6 a b c d+21 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{c+d x^4} \, dx,x,\sqrt {x}\right )}{16 c^2 d^2}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{5/2} d^2}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{5/2} d^2}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{5/2} d^{5/2}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{5/2} d^{5/2}}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}\\ &=\frac {(b c-a d)^2 \sqrt {x}}{4 c d^2 \left (c+d x^2\right )^2}-\frac {(b c-a d) (9 b c+7 a d) \sqrt {x}}{16 c^2 d^2 \left (c+d x^2\right )}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}-\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}+\frac {\left (5 b^2 c^2+6 a b c d+21 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^{11/4} d^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 339, normalized size = 0.93 \[ \frac {-\sqrt {2} \left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )+\sqrt {2} \left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )-2 \sqrt {2} \left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )+2 \sqrt {2} \left (21 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )-\frac {8 c^{3/4} \sqrt [4]{d} \sqrt {x} \left (-7 a^2 d^2-2 a b c d+9 b^2 c^2\right )}{c+d x^2}+\frac {32 c^{7/4} \sqrt [4]{d} \sqrt {x} (b c-a d)^2}{\left (c+d x^2\right )^2}}{128 c^{11/4} d^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((32*c^(7/4)*d^(1/4)*(b*c - a*d)^2*Sqrt[x])/(c + d*x^2)^2 - (8*c^(3/4)*d^(1/4)*(9*b^2*c^2 - 2*a*b*c*d - 7*a^2*
d^2)*Sqrt[x])/(c + d*x^2) - 2*Sqrt[2]*(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x]
)/c^(1/4)] + 2*Sqrt[2]*(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)] - Sq
rt[2]*(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x] + Sqrt[2
]*(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(128*c^(11/
4)*d^(9/4))

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fricas [B]  time = 0.51, size = 1416, normalized size = 3.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^3/x^(1/2),x, algorithm="fricas")

[Out]

1/64*(4*(c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)*(-(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42
120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*
c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(1/4)*arctan((sqrt(c^6*d^4*sqrt(-(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a
^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*
d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9)) + (25*b^4*c^4 + 60*a*b^3*c^3*d + 246*a^2*b^2*c^2*d^2 +
252*a^3*b*c*d^3 + 441*a^4*d^4)*x)*c^8*d^7*(-(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^
3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7
+ 194481*a^8*d^8)/(c^11*d^9))^(3/4) - (5*b^2*c^10*d^7 + 6*a*b*c^9*d^8 + 21*a^2*c^8*d^9)*sqrt(x)*(-(625*b^8*c^8
+ 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*
c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(3/4))/(625*b^8*c^8 + 3000
*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5
+ 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)) + (c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)*(-
(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176
904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(1/4)*log(c^3*
d^2*(-(625*b^8*c^8 + 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4
+ 176904*a^5*b^3*c^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(1/4) +
(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*sqrt(x)) - (c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)*(-(625*b^8*c^8 + 3000*
a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c^3*d^5
+ 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(1/4)*log(-c^3*d^2*(-(625*b^8*c^8
+ 3000*a*b^7*c^7*d + 15900*a^2*b^6*c^6*d^2 + 42120*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 176904*a^5*b^3*c
^3*d^5 + 280476*a^6*b^2*c^2*d^6 + 222264*a^7*b*c*d^7 + 194481*a^8*d^8)/(c^11*d^9))^(1/4) + (5*b^2*c^2 + 6*a*b*
c*d + 21*a^2*d^2)*sqrt(x)) - 4*(5*b^2*c^3 + 6*a*b*c^2*d - 11*a^2*c*d^2 + (9*b^2*c^2*d - 2*a*b*c*d^2 - 7*a^2*d^
3)*x^2)*sqrt(x))/(c^2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2)

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giac [A]  time = 0.53, size = 416, normalized size = 1.14 \[ \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 21 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{3}} + \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 21 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{64 \, c^{3} d^{3}} + \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 21 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{3}} - \frac {\sqrt {2} {\left (5 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 6 \, \left (c d^{3}\right )^{\frac {1}{4}} a b c d + 21 \, \left (c d^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{128 \, c^{3} d^{3}} - \frac {9 \, b^{2} c^{2} d x^{\frac {5}{2}} - 2 \, a b c d^{2} x^{\frac {5}{2}} - 7 \, a^{2} d^{3} x^{\frac {5}{2}} + 5 \, b^{2} c^{3} \sqrt {x} + 6 \, a b c^{2} d \sqrt {x} - 11 \, a^{2} c d^{2} \sqrt {x}}{16 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^3/x^(1/2),x, algorithm="giac")

[Out]

1/64*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^(1/4)*a*b*c*d + 21*(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^3*d^3) + 1/64*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^
(1/4)*a*b*c*d + 21*(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^3*d^3) + 1/128*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^(1/4)*a*b*c*d + 21*(c*d^3)^(1/4)*a^2*d^2)*log(sq
rt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^3) - 1/128*sqrt(2)*(5*(c*d^3)^(1/4)*b^2*c^2 + 6*(c*d^3)^(1/4
)*a*b*c*d + 21*(c*d^3)^(1/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^3) - 1/16*(9*b^
2*c^2*d*x^(5/2) - 2*a*b*c*d^2*x^(5/2) - 7*a^2*d^3*x^(5/2) + 5*b^2*c^3*sqrt(x) + 6*a*b*c^2*d*sqrt(x) - 11*a^2*c
*d^2*sqrt(x))/((d*x^2 + c)^2*c^2*d^2)

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maple [A]  time = 0.02, size = 514, normalized size = 1.41 \[ \frac {21 \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^{3}}+\frac {21 \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^{3}}+\frac {21 \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^{3}}+\frac {3 \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^{2} d}+\frac {3 \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^{2} d}+\frac {3 \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^{2} d}+\frac {5 \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 c \,d^{2}}+\frac {5 \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 c \,d^{2}}+\frac {5 \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 c \,d^{2}}+\frac {\frac {\left (7 a^{2} d^{2}+2 a b c d -9 b^{2} c^{2}\right ) x^{\frac {5}{2}}}{16 c^{2} d}+\frac {\left (11 a^{2} d^{2}-6 a b c d -5 b^{2} c^{2}\right ) \sqrt {x}}{16 c \,d^{2}}}{\left (d \,x^{2}+c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(d*x^2+c)^3/x^(1/2),x)

[Out]

2*(1/32*(7*a^2*d^2+2*a*b*c*d-9*b^2*c^2)/c^2/d*x^(5/2)+1/32*(11*a^2*d^2-6*a*b*c*d-5*b^2*c^2)/c/d^2*x^(1/2))/(d*
x^2+c)^2+21/64/c^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+3/32/c^2/d*(c/d)^(1/4)*2^(1/2
)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b+5/64/c/d^2*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-
1)*b^2+21/128/c^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)))*a^2+3/64/c^2/d*(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+5/128/c/d^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)))*b^2+21/64/c^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^
(1/4)*x^(1/2)+1)*a^2+3/32/c^2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b+5/64/c/d^2*(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 = 2.43, size = 366, normalized size = 1.01 \[ -\frac {{\left (9 \, b^{2} c^{2} d - 2 \, a b c d^{2} - 7 \, a^{2} d^{3}\right )} x^{\frac {5}{2}} + {\left (5 \, b^{2} c^{3} + 6 \, a b c^{2} d - 11 \, a^{2} c d^{2}\right )} \sqrt {x}}{16 \, {\left (c^{2} d^{4} x^{4} + 2 \, c^{3} d^{3} x^{2} + c^{4} d^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (5 \, b^{2} c^{2} + 6 \, a b c d + 21 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (5 \, b^{2} c^{2} + 6 \, a b c d + 21 \, a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (5 \, b^{2} c^{2} + 6 \, a b c d + 21 \, a^{2} d^{2}\right )} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, b^{2} c^{2} + 6 \, a b c d + 21 \, a^{2} d^{2}\right )} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{128 \, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^3/x^(1/2),x, algorithm="maxima")

[Out]

-1/16*((9*b^2*c^2*d - 2*a*b*c*d^2 - 7*a^2*d^3)*x^(5/2) + (5*b^2*c^3 + 6*a*b*c^2*d - 11*a^2*c*d^2)*sqrt(x))/(c^
2*d^4*x^4 + 2*c^3*d^3*x^2 + c^4*d^2) + 1/128*(2*sqrt(2)*(5*b^2*c^2 + 6*a*b*c*d + 21*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*sqr
t(2)*(5*b^2*c^2 + 6*a*b*c*d + 21*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2*sqrt(d)*sqrt(x))/sq
rt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(5*b^2*c^2 + 6*a*b*c*d + 21*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)*(5*b^2*c^2 + 6*a*b*c*d + 21*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^2*d^2)

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mupad [B]  time = 0.61, size = 1419, normalized size = 3.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^(1/2)*(c + d*x^2)^3),x)

[Out]

- ((x^(1/2)*(5*b^2*c^2 - 11*a^2*d^2 + 6*a*b*c*d))/(16*c*d^2) - (x^(5/2)*(7*a^2*d^2 - 9*b^2*c^2 + 2*a*b*c*d))/(
16*c^2*d))/(c^2 + d^2*x^4 + 2*c*d*x^2) - (atan((((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^
2*d + 6*a*b*c*d^2))/(64*(-c)^(15/4)*d^(9/4)) - (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a
*b^3*c^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*1i)/(64*(-c)^(11/4)*d^(9/4)) -
((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^2*d + 6*a*b*c*d^2))/(64*(-c)^(15/4)*d^(9/4)) +
(x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*
a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*1i)/(64*(-c)^(11/4)*d^(9/4)))/(((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^
2*d^3 + 5*b^2*c^2*d + 6*a*b*c*d^2))/(64*(-c)^(15/4)*d^(9/4)) - (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^
2*c^2*d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d))/(64*(-c)^(11/
4)*d^(9/4)) + ((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^2*d + 6*a*b*c*d^2))/(64*(-c)^(15/4
)*d^(9/4)) + (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64
*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d))/(64*(-c)^(11/4)*d^(9/4))))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)
*1i)/(32*(-c)^(11/4)*d^(9/4)) - (atan((((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^2*d + 6*a
*b*c*d^2)*1i)/(64*(-c)^(15/4)*d^(9/4)) - (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a*b^3*c
^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d))/(64*(-c)^(11/4)*d^(9/4)) - ((((21*a
^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^2*d + 6*a*b*c*d^2)*1i)/(64*(-c)^(15/4)*d^(9/4)) + (x^(1/
2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*a^2*d^
2 + 5*b^2*c^2 + 6*a*b*c*d))/(64*(-c)^(11/4)*d^(9/4)))/(((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5
*b^2*c^2*d + 6*a*b*c*d^2)*1i)/(64*(-c)^(15/4)*d^(9/4)) - (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*
d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*1i)/(64*(-c)^(11/4)*
d^(9/4)) + ((((21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*(21*a^2*d^3 + 5*b^2*c^2*d + 6*a*b*c*d^2)*1i)/(64*(-c)^(15/4
)*d^(9/4)) + (x^(1/2)*(441*a^4*d^4 + 25*b^4*c^4 + 246*a^2*b^2*c^2*d^2 + 60*a*b^3*c^3*d + 252*a^3*b*c*d^3))/(64
*c^4*d))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c*d)*1i)/(64*(-c)^(11/4)*d^(9/4))))*(21*a^2*d^2 + 5*b^2*c^2 + 6*a*b*c
*d))/(32*(-c)^(11/4)*d^(9/4))

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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((b*x**2+a)**2/(d*x**2+c)**3/x**(1/2),x)

[Out]

Timed out

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