3.486 \(\int \frac{1}{x^{5/2} \left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx\)

Optimal. Leaf size=681 \[ \frac{b^{15/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} (b c-a d)^3}-\frac{b^{15/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} (b c-a d)^3}+\frac{b^{15/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} (b c-a d)^3}-\frac{b^{15/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{7/4} (b c-a d)^3}-\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} (b c-a d)^3}+\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} (b c-a d)^3}-\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{32 \sqrt{2} c^{15/4} (b c-a d)^3}+\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt{2} c^{15/4} (b c-a d)^3}-\frac{77 a^2 d^2-133 a b c d+32 b^2 c^2}{48 a c^3 x^{3/2} (b c-a d)^2}-\frac{d (19 b c-11 a d)}{16 c^2 x^{3/2} \left (c+d x^2\right ) (b c-a d)^2}-\frac{d}{4 c x^{3/2} \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

-(32*b^2*c^2 - 133*a*b*c*d + 77*a^2*d^2)/(48*a*c^3*(b*c - a*d)^2*x^(3/2)) - d/(4
*c*(b*c - a*d)*x^(3/2)*(c + d*x^2)^2) - (d*(19*b*c - 11*a*d))/(16*c^2*(b*c - a*d
)^2*x^(3/2)*(c + d*x^2)) + (b^(15/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4
)])/(Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (b^(15/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (d^(7/4)*(165*b^2*c^2 - 210*a*b
*c*d + 77*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^
(15/4)*(b*c - a*d)^3) + (d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*d^2)*ArcTan
[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(15/4)*(b*c - a*d)^3) + (
b^(15/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*
a^(7/4)*(b*c - a*d)^3) - (b^(15/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]
+ Sqrt[b]*x])/(2*Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (d^(7/4)*(165*b^2*c^2 - 210*a
*b*c*d + 77*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])
/(64*Sqrt[2]*c^(15/4)*(b*c - a*d)^3) + (d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*
a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]
*c^(15/4)*(b*c - a*d)^3)

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Rubi [A]  time = 2.01855, antiderivative size = 681, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 11, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.458 \[ \frac{b^{15/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} (b c-a d)^3}-\frac{b^{15/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{2 \sqrt{2} a^{7/4} (b c-a d)^3}+\frac{b^{15/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt{2} a^{7/4} (b c-a d)^3}-\frac{b^{15/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt{2} a^{7/4} (b c-a d)^3}-\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} (b c-a d)^3}+\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} (b c-a d)^3}-\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{32 \sqrt{2} c^{15/4} (b c-a d)^3}+\frac{d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt{2} c^{15/4} (b c-a d)^3}-\frac{77 a^2 d^2-133 a b c d+32 b^2 c^2}{48 a c^3 x^{3/2} (b c-a d)^2}-\frac{d (19 b c-11 a d)}{16 c^2 x^{3/2} \left (c+d x^2\right ) (b c-a d)^2}-\frac{d}{4 c x^{3/2} \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]  Int[1/(x^(5/2)*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

-(32*b^2*c^2 - 133*a*b*c*d + 77*a^2*d^2)/(48*a*c^3*(b*c - a*d)^2*x^(3/2)) - d/(4
*c*(b*c - a*d)*x^(3/2)*(c + d*x^2)^2) - (d*(19*b*c - 11*a*d))/(16*c^2*(b*c - a*d
)^2*x^(3/2)*(c + d*x^2)) + (b^(15/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4
)])/(Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (b^(15/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[x])/a^(1/4)])/(Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (d^(7/4)*(165*b^2*c^2 - 210*a*b
*c*d + 77*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^
(15/4)*(b*c - a*d)^3) + (d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*d^2)*ArcTan
[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(15/4)*(b*c - a*d)^3) + (
b^(15/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*
a^(7/4)*(b*c - a*d)^3) - (b^(15/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]
+ Sqrt[b]*x])/(2*Sqrt[2]*a^(7/4)*(b*c - a*d)^3) - (d^(7/4)*(165*b^2*c^2 - 210*a
*b*c*d + 77*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])
/(64*Sqrt[2]*c^(15/4)*(b*c - a*d)^3) + (d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*
a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]
*c^(15/4)*(b*c - a*d)^3)

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/x**(5/2)/(b*x**2+a)/(d*x**2+c)**3,x)

[Out]

Timed out

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Mathematica [A]  time = 2.61724, size = 639, normalized size = 0.94 \[ \frac{1}{384} \left (\frac{96 \sqrt{2} b^{15/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{7/4} (b c-a d)^3}+\frac{96 \sqrt{2} b^{15/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{7/4} (a d-b c)^3}-\frac{192 \sqrt{2} b^{15/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{a^{7/4} (a d-b c)^3}+\frac{192 \sqrt{2} b^{15/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{a^{7/4} (a d-b c)^3}+\frac{3 \sqrt{2} d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{15/4} (a d-b c)^3}+\frac{3 \sqrt{2} d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{c^{15/4} (b c-a d)^3}-\frac{6 \sqrt{2} d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{c^{15/4} (b c-a d)^3}+\frac{6 \sqrt{2} d^{7/4} \left (77 a^2 d^2-210 a b c d+165 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{c^{15/4} (b c-a d)^3}+\frac{24 d^2 \sqrt{x} (23 b c-15 a d)}{c^3 \left (c+d x^2\right ) (b c-a d)^2}+\frac{96 d^2 \sqrt{x}}{c^2 \left (c+d x^2\right )^2 (b c-a d)}-\frac{256}{a c^3 x^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[1/(x^(5/2)*(a + b*x^2)*(c + d*x^2)^3),x]

[Out]

(-256/(a*c^3*x^(3/2)) + (96*d^2*Sqrt[x])/(c^2*(b*c - a*d)*(c + d*x^2)^2) + (24*d
^2*(23*b*c - 15*a*d)*Sqrt[x])/(c^3*(b*c - a*d)^2*(c + d*x^2)) - (192*Sqrt[2]*b^(
15/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(a^(7/4)*(-(b*c) + a*d)^3)
+ (192*Sqrt[2]*b^(15/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(a^(7/4)*
(-(b*c) + a*d)^3) - (6*Sqrt[2]*d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*d^2)*
ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(c^(15/4)*(b*c - a*d)^3) + (6*Sqr
t[2]*d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4
)*Sqrt[x])/c^(1/4)])/(c^(15/4)*(b*c - a*d)^3) + (96*Sqrt[2]*b^(15/4)*Log[Sqrt[a]
- Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(7/4)*(b*c - a*d)^3) + (96*S
qrt[2]*b^(15/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(a^(
7/4)*(-(b*c) + a*d)^3) + (3*Sqrt[2]*d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*
d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(c^(15/4)*(-(b*
c) + a*d)^3) + (3*Sqrt[2]*d^(7/4)*(165*b^2*c^2 - 210*a*b*c*d + 77*a^2*d^2)*Log[S
qrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(c^(15/4)*(b*c - a*d)^3))
/384

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Maple [A]  time = 0.035, size = 906, normalized size = 1.3 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/x^(5/2)/(b*x^2+a)/(d*x^2+c)^3,x)

[Out]

-15/16*d^5/c^3/(a*d-b*c)^3/(d*x^2+c)^2*x^(5/2)*a^2+19/8*d^4/c^2/(a*d-b*c)^3/(d*x
^2+c)^2*x^(5/2)*a*b-23/16*d^3/c/(a*d-b*c)^3/(d*x^2+c)^2*x^(5/2)*b^2-19/16*d^4/c^
2/(a*d-b*c)^3/(d*x^2+c)^2*x^(1/2)*a^2+23/8*d^3/c/(a*d-b*c)^3/(d*x^2+c)^2*x^(1/2)
*a*b-27/16*d^2/(a*d-b*c)^3/(d*x^2+c)^2*x^(1/2)*b^2-77/64*d^4/c^4/(a*d-b*c)^3*(c/
d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a^2+105/32*d^3/c^3/(a*d-b
*c)^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*a*b-165/64*d^2/c
^2/(a*d-b*c)^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-77/
128*d^4/c^4/(a*d-b*c)^3*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c
/d)^(1/2))/(x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))*a^2+105/64*d^3/c^3/(a*d-
b*c)^3*(c/d)^(1/4)*2^(1/2)*ln((x+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2))/(x-(c/
d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))*a*b-165/128*d^2/c^2/(a*d-b*c)^3*(c/d)^(1/
4)*2^(1/2)*ln((x+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2))/(x-(c/d)^(1/4)*x^(1/2)
*2^(1/2)+(c/d)^(1/2)))*b^2-77/64*d^4/c^4/(a*d-b*c)^3*(c/d)^(1/4)*2^(1/2)*arctan(
2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+105/32*d^3/c^3/(a*d-b*c)^3*(c/d)^(1/4)*2^(1/2
)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a*b-165/64*d^2/c^2/(a*d-b*c)^3*(c/d)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*b^2-2/3/a/c^3/x^(3/2)+1/4/a^2*b
^4/(a*d-b*c)^3*(a/b)^(1/4)*2^(1/2)*ln((x+(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)
)/(x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+1/2/a^2*b^4/(a*d-b*c)^3*(a/b)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+1/2/a^2*b^4/(a*d-b*c)^3*(a/b)^(
1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((b*x^2 + a)*(d*x^2 + c)^3*x^(5/2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((b*x^2 + a)*(d*x^2 + c)^3*x^(5/2)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/x**(5/2)/(b*x**2+a)/(d*x**2+c)**3,x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.402415, size = 1, normalized size = 0. \[ \mathit{Done} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((b*x^2 + a)*(d*x^2 + c)^3*x^(5/2)),x, algorithm="giac")

[Out]

Done