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

Optimal. Leaf size=333 \[ -\frac {x^{3/2} \left (5 a^2 d^2-2 a b c d+b^2 c^2\right )}{2 c^2 d \left (c+d x^2\right )}-\frac {2 a^2}{c \sqrt {x} \left (c+d x^2\right )}+\frac {(b c-a d) (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{9/4} d^{7/4}}-\frac {(b c-a d) (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{9/4} d^{7/4}}-\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{9/4} d^{7/4}}+\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{9/4} d^{7/4}} \]

[Out]

-1/2*(5*a^2*d^2-2*a*b*c*d+b^2*c^2)*x^(3/2)/c^2/d/(d*x^2+c)-1/8*(-a*d+b*c)*(5*a*d+3*b*c)*arctan(1-d^(1/4)*2^(1/
2)*x^(1/2)/c^(1/4))/c^(9/4)/d^(7/4)*2^(1/2)+1/8*(-a*d+b*c)*(5*a*d+3*b*c)*arctan(1+d^(1/4)*2^(1/2)*x^(1/2)/c^(1
/4))/c^(9/4)/d^(7/4)*2^(1/2)+1/16*(-a*d+b*c)*(5*a*d+3*b*c)*ln(c^(1/2)+x*d^(1/2)-c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2
))/c^(9/4)/d^(7/4)*2^(1/2)-1/16*(-a*d+b*c)*(5*a*d+3*b*c)*ln(c^(1/2)+x*d^(1/2)+c^(1/4)*d^(1/4)*2^(1/2)*x^(1/2))
/c^(9/4)/d^(7/4)*2^(1/2)-2*a^2/c/(d*x^2+c)/x^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 333, 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 = {462, 457, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {x^{3/2} \left (5 a^2 d^2-2 a b c d+b^2 c^2\right )}{2 c^2 d \left (c+d x^2\right )}-\frac {2 a^2}{c \sqrt {x} \left (c+d x^2\right )}+\frac {(b c-a d) (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{9/4} d^{7/4}}-\frac {(b c-a d) (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} c^{9/4} d^{7/4}}-\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} c^{9/4} d^{7/4}}+\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} c^{9/4} d^{7/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(c*Sqrt[x]*(c + d*x^2)) - ((b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x^(3/2))/(2*c^2*d*(c + d*x^2)) - ((b*c -
a*d)*(3*b*c + 5*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(9/4)*d^(7/4)) + ((b*c - a*d
)*(3*b*c + 5*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(9/4)*d^(7/4)) + ((b*c - a*d)*(3
*b*c + 5*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(9/4)*d^(7/4)) - ((b*c
- a*d)*(3*b*c + 5*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[2]*c^(9/4)*d^(7/4))

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 462

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

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

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Mathematica [A]  time = 0.20, size = 317, normalized size = 0.95 \[ \frac {\frac {\sqrt {2} \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{d^{7/4}}+\frac {\sqrt {2} \left (5 a^2 d^2-2 a b c d-3 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{d^{7/4}}+\frac {2 \sqrt {2} \left (5 a^2 d^2-2 a b c d-3 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{d^{7/4}}+\frac {2 \sqrt {2} \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{d^{7/4}}-\frac {32 a^2 \sqrt [4]{c}}{\sqrt {x}}-\frac {8 \sqrt [4]{c} x^{3/2} (b c-a d)^2}{d \left (c+d x^2\right )}}{16 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*a^2*c^(1/4))/Sqrt[x] - (8*c^(1/4)*(b*c - a*d)^2*x^(3/2))/(d*(c + d*x^2)) + (2*Sqrt[2]*(-3*b^2*c^2 - 2*a*
b*c*d + 5*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(7/4) + (2*Sqrt[2]*(3*b^2*c^2 + 2*a*b*c*d
- 5*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(7/4) + (Sqrt[2]*(3*b^2*c^2 + 2*a*b*c*d - 5*a^2*
d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(7/4) + (Sqrt[2]*(-3*b^2*c^2 - 2*a*b*c*d +
5*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/d^(7/4))/(16*c^(9/4))

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fricas [B]  time = 0.63, size = 1739, normalized size = 5.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(c^2*d^2*x^3 + c^3*d*x)*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 6
46*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7))^(
1/4)*arctan((sqrt((729*b^12*c^12 + 2916*a*b^11*c^11*d - 2430*a^2*b^10*c^10*d^2 - 19980*a^3*b^9*c^9*d^3 + 135*a
^4*b^8*c^8*d^4 + 59976*a^5*b^7*c^7*d^5 + 6364*a^6*b^6*c^6*d^6 - 99960*a^7*b^5*c^5*d^7 + 375*a^8*b^4*c^4*d^8 +
92500*a^9*b^3*c^3*d^9 - 18750*a^10*b^2*c^2*d^10 - 37500*a^11*b*c*d^11 + 15625*a^12*d^12)*x - (81*b^8*c^13*d^3
+ 216*a*b^7*c^12*d^4 - 324*a^2*b^6*c^11*d^5 - 984*a^3*b^5*c^10*d^6 + 646*a^4*b^4*c^9*d^7 + 1640*a^5*b^3*c^8*d^
8 - 900*a^6*b^2*c^7*d^9 - 1000*a^7*b*c^6*d^10 + 625*a^8*c^5*d^11)*sqrt(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^
2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*
a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7)))*c^2*d^2*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3
*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d
^8)/(c^9*d^7))^(1/4) + (27*b^6*c^8*d^2 + 54*a*b^5*c^7*d^3 - 99*a^2*b^4*c^6*d^4 - 172*a^3*b^3*c^5*d^5 + 165*a^4
*b^2*c^4*d^6 + 150*a^5*b*c^3*d^7 - 125*a^6*c^2*d^8)*sqrt(x)*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*
d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^
7 + 625*a^8*d^8)/(c^9*d^7))^(1/4))/(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 +
646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)) - (c^2*d^
2*x^3 + c^3*d*x)*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4
*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7))^(1/4)*log(c^7*d
^5*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^
5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7))^(3/4) - (27*b^6*c^6 + 54*a*b^
5*c^5*d - 99*a^2*b^4*c^4*d^2 - 172*a^3*b^3*c^3*d^3 + 165*a^4*b^2*c^2*d^4 + 150*a^5*b*c*d^5 - 125*a^6*d^6)*sqrt
(x)) + (c^2*d^2*x^3 + c^3*d*x)*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 6
46*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7))^(
1/4)*log(-c^7*d^5*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^
4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(c^9*d^7))^(3/4) - (27*b^
6*c^6 + 54*a*b^5*c^5*d - 99*a^2*b^4*c^4*d^2 - 172*a^3*b^3*c^3*d^3 + 165*a^4*b^2*c^2*d^4 + 150*a^5*b*c*d^5 - 12
5*a^6*d^6)*sqrt(x)) - 4*(4*a^2*c*d + (b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x^2)*sqrt(x))/(c^2*d^2*x^3 + c^3*d*x)

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giac [A]  time = 0.53, size = 389, normalized size = 1.17 \[ -\frac {b^{2} c^{2} x^{2} - 2 \, a b c d x^{2} + 5 \, a^{2} d^{2} x^{2} + 4 \, a^{2} c d}{2 \, {\left (d x^{\frac {5}{2}} + c \sqrt {x}\right )} c^{2} d} + \frac {\sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d - 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3} d^{4}} + \frac {\sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d - 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, c^{3} d^{4}} - \frac {\sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d - 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, c^{3} d^{4}} + \frac {\sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} + 2 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d - 5 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, c^{3} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*c^2*x^2 - 2*a*b*c*d*x^2 + 5*a^2*d^2*x^2 + 4*a^2*c*d)/((d*x^(5/2) + c*sqrt(x))*c^2*d) + 1/8*sqrt(2)*(
3*(c*d^3)^(3/4)*b^2*c^2 + 2*(c*d^3)^(3/4)*a*b*c*d - 5*(c*d^3)^(3/4)*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^4) + 1/8*sqrt(2)*(3*(c*d^3)^(3/4)*b^2*c^2 + 2*(c*d^3)^(3/4)*a*b*c*d -
5*(c*d^3)^(3/4)*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^4) - 1/16*s
qrt(2)*(3*(c*d^3)^(3/4)*b^2*c^2 + 2*(c*d^3)^(3/4)*a*b*c*d - 5*(c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)
^(1/4) + x + sqrt(c/d))/(c^3*d^4) + 1/16*sqrt(2)*(3*(c*d^3)^(3/4)*b^2*c^2 + 2*(c*d^3)^(3/4)*a*b*c*d - 5*(c*d^3
)^(3/4)*a^2*d^2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c^3*d^4)

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maple [A]  time = 0.02, size = 495, normalized size = 1.49 \[ -\frac {a^{2} d \,x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) c^{2}}+\frac {a b \,x^{\frac {3}{2}}}{\left (d \,x^{2}+c \right ) c}-\frac {b^{2} x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) d}-\frac {5 \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}-\frac {5 \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}-\frac {5 \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 )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} c^{2}}+\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} c d}+\frac {\sqrt {2}\, a b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} c d}+\frac {\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 )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} c d}+\frac {3 \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \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 )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {2 a^{2}}{c^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2*d*x^(3/2)/(d*x^2+c)*a^2+1/c*x^(3/2)/(d*x^2+c)*a*b-1/2/d*x^(3/2)/(d*x^2+c)*b^2-5/8/c^2/(c/d)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+1/4/c/d/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-
1)*a*b+3/8/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-5/16/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+1/8/c/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+3/16
/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^2-5/8/c^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+1/4/c/d/(c/d)^(1/4)*2^(1/2)*a
rctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b+3/8/d^2/(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2
-2*a^2/c^2/x^(1/2)

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maxima [A]  time = 2.38, size = 260, normalized size = 0.78 \[ -\frac {4 \, a^{2} c d + {\left (b^{2} c^{2} - 2 \, a b c d + 5 \, a^{2} d^{2}\right )} x^{2}}{2 \, {\left (c^{2} d^{2} x^{\frac {5}{2}} + c^{3} d \sqrt {x}\right )}} + \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d - 5 \, a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{16 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a^2*c*d + (b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x^2)/(c^2*d^2*x^(5/2) + c^3*d*sqrt(x)) + 1/16*(3*b^2*c^2 +
2*a*b*c*d - 5*a^2*d^2)*(2*sqrt(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(sqrt(c)*sqrt(d))*sqrt(d)) + 2*sqrt(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(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt
(x) + sqrt(d)*x + sqrt(c))/(c^(1/4)*d^(3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt
(c))/(c^(1/4)*d^(3/4)))/(c^2*d)

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mupad [B]  time = 0.39, size = 138, normalized size = 0.41 \[ \frac {\mathrm {atanh}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a\,d+3\,b\,c\right )}{4\,{\left (-c\right )}^{9/4}\,d^{7/4}}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (5\,a\,d+3\,b\,c\right )}{4\,{\left (-c\right )}^{9/4}\,d^{7/4}}-\frac {\frac {2\,a^2}{c}+\frac {x^2\,\left (5\,a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,c^2\,d}}{c\,\sqrt {x}+d\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((d^(1/4)*x^(1/2))/(-c)^(1/4))*(a*d - b*c)*(5*a*d + 3*b*c))/(4*(-c)^(9/4)*d^(7/4)) - (atan((d^(1/4)*x^(1
/2))/(-c)^(1/4))*(a*d - b*c)*(5*a*d + 3*b*c))/(4*(-c)^(9/4)*d^(7/4)) - ((2*a^2)/c + (x^2*(5*a^2*d^2 + b^2*c^2
- 2*a*b*c*d))/(2*c^2*d))/(c*x^(1/2) + d*x^(5/2))

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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/x**(3/2)/(d*x**2+c)**2,x)

[Out]

Timed out

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