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

Optimal. Leaf size=402 \[ -\frac{\sqrt{x} \left (11 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{12 c^2 d \left (c+d x^2\right )^2}-\frac{2 a^2}{3 c x^{3/2} \left (c+d x^2\right )^2}-\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} d^{5/4}}+\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} d^{5/4}}-\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{32 \sqrt{2} c^{15/4} d^{5/4}}+\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt{2} c^{15/4} d^{5/4}}+\frac{\sqrt{x} \left (7 a d (6 b c-11 a d)+3 b^2 c^2\right )}{48 c^3 d \left (c+d x^2\right )} \]

[Out]

(-2*a^2)/(3*c*x^(3/2)*(c + d*x^2)^2) - ((3*b^2*c^2 - 6*a*b*c*d + 11*a^2*d^2)*Sqr
t[x])/(12*c^2*d*(c + d*x^2)^2) + ((3*b^2*c^2 + 7*a*d*(6*b*c - 11*a*d))*Sqrt[x])/
(48*c^3*d*(c + d*x^2)) - ((3*b^2*c^2 + 7*a*d*(6*b*c - 11*a*d))*ArcTan[1 - (Sqrt[
2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(15/4)*d^(5/4)) + ((3*b^2*c^2 + 7*a*
d*(6*b*c - 11*a*d))*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c
^(15/4)*d^(5/4)) - ((3*b^2*c^2 + 7*a*d*(6*b*c - 11*a*d))*Log[Sqrt[c] - Sqrt[2]*c
^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(15/4)*d^(5/4)) + ((3*b^2*c^2
+ 7*a*d*(6*b*c - 11*a*d))*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[
d]*x])/(64*Sqrt[2]*c^(15/4)*d^(5/4))

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Rubi [A]  time = 0.845902, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417 \[ -\frac{\sqrt{x} \left (11 a^2 d^2-6 a b c d+3 b^2 c^2\right )}{12 c^2 d \left (c+d x^2\right )^2}-\frac{2 a^2}{3 c x^{3/2} \left (c+d x^2\right )^2}+\frac{\sqrt{x} \left (\frac{7 a (6 b c-11 a d)}{c^2}+\frac{3 b^2}{d}\right )}{48 c \left (c+d x^2\right )}-\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} d^{5/4}}+\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{64 \sqrt{2} c^{15/4} d^{5/4}}-\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{32 \sqrt{2} c^{15/4} d^{5/4}}+\frac{\left (7 a d (6 b c-11 a d)+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{32 \sqrt{2} c^{15/4} d^{5/4}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*a^2)/(3*c*x^(3/2)*(c + d*x^2)^2) - ((3*b^2*c^2 - 6*a*b*c*d + 11*a^2*d^2)*Sqr
t[x])/(12*c^2*d*(c + d*x^2)^2) + (((3*b^2)/d + (7*a*(6*b*c - 11*a*d))/c^2)*Sqrt[
x])/(48*c*(c + d*x^2)) - ((3*b^2*c^2 + 7*a*d*(6*b*c - 11*a*d))*ArcTan[1 - (Sqrt[
2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c^(15/4)*d^(5/4)) + ((3*b^2*c^2 + 7*a*
d*(6*b*c - 11*a*d))*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(32*Sqrt[2]*c
^(15/4)*d^(5/4)) - ((3*b^2*c^2 + 7*a*d*(6*b*c - 11*a*d))*Log[Sqrt[c] - Sqrt[2]*c
^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(64*Sqrt[2]*c^(15/4)*d^(5/4)) + ((3*b^2*c^2
+ 7*a*d*(6*b*c - 11*a*d))*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[
d]*x])/(64*Sqrt[2]*c^(15/4)*d^(5/4))

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Rubi in Sympy [A]  time = 106.729, size = 384, normalized size = 0.96 \[ - \frac{2 a^{2}}{3 c x^{\frac{3}{2}} \left (c + d x^{2}\right )^{2}} - \frac{\sqrt{x} \left (a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right )}{12 c^{2} d \left (c + d x^{2}\right )^{2}} + \frac{\sqrt{x} \left (- 7 a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right )}{48 c^{3} d \left (c + d x^{2}\right )} - \frac{\sqrt{2} \left (- 7 a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right ) \log{\left (- \sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x} + \sqrt{c} + \sqrt{d} x \right )}}{128 c^{\frac{15}{4}} d^{\frac{5}{4}}} + \frac{\sqrt{2} \left (- 7 a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right ) \log{\left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x} + \sqrt{c} + \sqrt{d} x \right )}}{128 c^{\frac{15}{4}} d^{\frac{5}{4}}} - \frac{\sqrt{2} \left (- 7 a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right ) \operatorname{atan}{\left (1 - \frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}} \right )}}{64 c^{\frac{15}{4}} d^{\frac{5}{4}}} + \frac{\sqrt{2} \left (- 7 a d \left (11 a d - 6 b c\right ) + 3 b^{2} c^{2}\right ) \operatorname{atan}{\left (1 + \frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}} \right )}}{64 c^{\frac{15}{4}} d^{\frac{5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*a**2/(3*c*x**(3/2)*(c + d*x**2)**2) - sqrt(x)*(a*d*(11*a*d - 6*b*c) + 3*b**2*
c**2)/(12*c**2*d*(c + d*x**2)**2) + sqrt(x)*(-7*a*d*(11*a*d - 6*b*c) + 3*b**2*c*
*2)/(48*c**3*d*(c + d*x**2)) - sqrt(2)*(-7*a*d*(11*a*d - 6*b*c) + 3*b**2*c**2)*l
og(-sqrt(2)*c**(1/4)*d**(1/4)*sqrt(x) + sqrt(c) + sqrt(d)*x)/(128*c**(15/4)*d**(
5/4)) + sqrt(2)*(-7*a*d*(11*a*d - 6*b*c) + 3*b**2*c**2)*log(sqrt(2)*c**(1/4)*d**
(1/4)*sqrt(x) + sqrt(c) + sqrt(d)*x)/(128*c**(15/4)*d**(5/4)) - sqrt(2)*(-7*a*d*
(11*a*d - 6*b*c) + 3*b**2*c**2)*atan(1 - sqrt(2)*d**(1/4)*sqrt(x)/c**(1/4))/(64*
c**(15/4)*d**(5/4)) + sqrt(2)*(-7*a*d*(11*a*d - 6*b*c) + 3*b**2*c**2)*atan(1 + s
qrt(2)*d**(1/4)*sqrt(x)/c**(1/4))/(64*c**(15/4)*d**(5/4))

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Mathematica [A]  time = 0.447418, size = 365, normalized size = 0.91 \[ \frac{\frac{3 \sqrt{2} \left (77 a^2 d^2-42 a b c d-3 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{5/4}}+\frac{3 \sqrt{2} \left (-77 a^2 d^2+42 a b c d+3 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} \sqrt{x}+\sqrt{c}+\sqrt{d} x\right )}{d^{5/4}}+\frac{6 \sqrt{2} \left (77 a^2 d^2-42 a b c d-3 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}\right )}{d^{5/4}}+\frac{6 \sqrt{2} \left (-77 a^2 d^2+42 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} \sqrt{x}}{\sqrt [4]{c}}+1\right )}{d^{5/4}}+\frac{24 c^{3/4} \sqrt{x} \left (-15 a^2 d^2+14 a b c d+b^2 c^2\right )}{d \left (c+d x^2\right )}-\frac{256 a^2 c^{3/4}}{x^{3/2}}-\frac{96 c^{7/4} \sqrt{x} (b c-a d)^2}{d \left (c+d x^2\right )^2}}{384 c^{15/4}} \]

Antiderivative was successfully verified.

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

[Out]

((-256*a^2*c^(3/4))/x^(3/2) - (96*c^(7/4)*(b*c - a*d)^2*Sqrt[x])/(d*(c + d*x^2)^
2) + (24*c^(3/4)*(b^2*c^2 + 14*a*b*c*d - 15*a^2*d^2)*Sqrt[x])/(d*(c + d*x^2)) +
(6*Sqrt[2]*(-3*b^2*c^2 - 42*a*b*c*d + 77*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sq
rt[x])/c^(1/4)])/d^(5/4) + (6*Sqrt[2]*(3*b^2*c^2 + 42*a*b*c*d - 77*a^2*d^2)*ArcT
an[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/d^(5/4) + (3*Sqrt[2]*(-3*b^2*c^2 - 42
*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
])/d^(5/4) + (3*Sqrt[2]*(3*b^2*c^2 + 42*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])/d^(5/4))/(384*c^(15/4))

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Maple [A]  time = 0.03, size = 562, normalized size = 1.4 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-15/16/c^3/(d*x^2+c)^2*x^(5/2)*a^2*d^2+7/8/c^2/(d*x^2+c)^2*x^(5/2)*a*b*d+1/16/c/
(d*x^2+c)^2*x^(5/2)*b^2-19/16/c^2/(d*x^2+c)^2*d*x^(1/2)*a^2+11/8/c/(d*x^2+c)^2*x
^(1/2)*a*b-3/16/(d*x^2+c)^2/d*x^(1/2)*b^2-77/64/c^4*d*(c/d)^(1/4)*2^(1/2)*arctan
(2^(1/2)/(c/d)^(1/4)*x^(1/2)+1)*a^2+21/32/c^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(c/d)^(1/4)*x^(1/2)+1)*a*b+3/64/c^2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^
(1/4)*x^(1/2)+1)*b^2-77/64/c^4*d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*
x^(1/2)-1)*a^2+21/32/c^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-
1)*a*b+3/64/c^2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)*b^2-
77/128/c^4*d*(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+21/64/c^3*(c/d)^(1/4)*2^(1/2)*l
n((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+3/128/c^2/d*(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^2-2/3*a^2/c^3/x^(3/
2)

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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((b*x^2 + a)^2/((d*x^2 + c)^3*x^(5/2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.272909, size = 1617, normalized size = 4.02 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/192*(128*a^2*c^2*d - 4*(3*b^2*c^2*d + 42*a*b*c*d^2 - 77*a^2*d^3)*x^4 + 4*(9*b
^2*c^3 - 66*a*b*c^2*d + 121*a^2*c*d^2)*x^2 - 12*(c^3*d^3*x^5 + 2*c^4*d^2*x^3 + c
^5*d*x)*sqrt(x)*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 + 53978
4*a^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 5727414
0*a^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5))^(1/4)*a
rctan(-c^4*d*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 + 539784*a
^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 57274140*a
^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5))^(1/4)/((3*
b^2*c^2 + 42*a*b*c*d - 77*a^2*d^2)*sqrt(x) - sqrt(c^8*d^2*sqrt(-(81*b^8*c^8 + 45
36*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 + 539784*a^3*b^5*c^5*d^3 - 1457946*a^4*b^
4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 57274140*a^6*b^2*c^2*d^6 - 76697544*a^7*b
*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5)) + (9*b^4*c^4 + 252*a*b^3*c^3*d + 1302*a^2
*b^2*c^2*d^2 - 6468*a^3*b*c*d^3 + 5929*a^4*d^4)*x))) + 3*(c^3*d^3*x^5 + 2*c^4*d^
2*x^3 + c^5*d*x)*sqrt(x)*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^
2 + 539784*a^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5
+ 57274140*a^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5)
)^(1/4)*log(c^4*d*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 + 539
784*a^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 57274
140*a^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5))^(1/4)
- (3*b^2*c^2 + 42*a*b*c*d - 77*a^2*d^2)*sqrt(x)) - 3*(c^3*d^3*x^5 + 2*c^4*d^2*x
^3 + c^5*d*x)*sqrt(x)*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 +
539784*a^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 5
7274140*a^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5))^(
1/4)*log(-c^4*d*(-(81*b^8*c^8 + 4536*a*b^7*c^7*d + 86940*a^2*b^6*c^6*d^2 + 53978
4*a^3*b^5*c^5*d^3 - 1457946*a^4*b^4*c^4*d^4 - 13854456*a^5*b^3*c^3*d^5 + 5727414
0*a^6*b^2*c^2*d^6 - 76697544*a^7*b*c*d^7 + 35153041*a^8*d^8)/(c^15*d^5))^(1/4) -
(3*b^2*c^2 + 42*a*b*c*d - 77*a^2*d^2)*sqrt(x)))/((c^3*d^3*x^5 + 2*c^4*d^2*x^3 +
c^5*d*x)*sqrt(x))

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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((b*x**2+a)**2/x**(5/2)/(d*x**2+c)**3,x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.302784, size = 575, normalized size = 1.43 \[ -\frac{2 \, a^{2}}{3 \, c^{3} x^{\frac{3}{2}}} + \frac{\sqrt{2}{\left (3 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} + 42 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 77 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{c}{d}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{c}{d}\right )^{\frac{1}{4}}}\right )}{64 \, c^{4} d^{2}} + \frac{\sqrt{2}{\left (3 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} + 42 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 77 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{c}{d}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{c}{d}\right )^{\frac{1}{4}}}\right )}{64 \, c^{4} d^{2}} + \frac{\sqrt{2}{\left (3 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} + 42 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 77 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\right )}{\rm ln}\left (\sqrt{2} \sqrt{x} \left (\frac{c}{d}\right )^{\frac{1}{4}} + x + \sqrt{\frac{c}{d}}\right )}{128 \, c^{4} d^{2}} - \frac{\sqrt{2}{\left (3 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} + 42 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 77 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\right )}{\rm ln}\left (-\sqrt{2} \sqrt{x} \left (\frac{c}{d}\right )^{\frac{1}{4}} + x + \sqrt{\frac{c}{d}}\right )}{128 \, c^{4} d^{2}} + \frac{b^{2} c^{2} d x^{\frac{5}{2}} + 14 \, a b c d^{2} x^{\frac{5}{2}} - 15 \, a^{2} d^{3} x^{\frac{5}{2}} - 3 \, b^{2} c^{3} \sqrt{x} + 22 \, a b c^{2} d \sqrt{x} - 19 \, a^{2} c d^{2} \sqrt{x}}{16 \,{\left (d x^{2} + c\right )}^{2} c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2/3*a^2/(c^3*x^(3/2)) + 1/64*sqrt(2)*(3*(c*d^3)^(1/4)*b^2*c^2 + 42*(c*d^3)^(1/4
)*a*b*c*d - 77*(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^4*d^2) + 1/64*sqrt(2)*(3*(c*d^3)^(1/4)*b^2*c^2 + 42*(
c*d^3)^(1/4)*a*b*c*d - 77*(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^4*d^2) + 1/128*sqrt(2)*(3*(c*d^3)^(1/4)*b
^2*c^2 + 42*(c*d^3)^(1/4)*a*b*c*d - 77*(c*d^3)^(1/4)*a^2*d^2)*ln(sqrt(2)*sqrt(x)
*(c/d)^(1/4) + x + sqrt(c/d))/(c^4*d^2) - 1/128*sqrt(2)*(3*(c*d^3)^(1/4)*b^2*c^2
+ 42*(c*d^3)^(1/4)*a*b*c*d - 77*(c*d^3)^(1/4)*a^2*d^2)*ln(-sqrt(2)*sqrt(x)*(c/d
)^(1/4) + x + sqrt(c/d))/(c^4*d^2) + 1/16*(b^2*c^2*d*x^(5/2) + 14*a*b*c*d^2*x^(5
/2) - 15*a^2*d^3*x^(5/2) - 3*b^2*c^3*sqrt(x) + 22*a*b*c^2*d*sqrt(x) - 19*a^2*c*d
^2*sqrt(x))/((d*x^2 + c)^2*c^3*d)