3.658 \(\int \frac{(a+b x)^{5/2} (c+d x)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=339 \[ -\frac{5 \sqrt{a+b x} (c+d x)^{5/2} \left (a^2 d^2+12 a b c d+3 b^2 c^2\right )}{24 c^2 x}+\frac{5 d \sqrt{a+b x} (c+d x)^{3/2} \left (a^2 d^2+14 a b c d+9 b^2 c^2\right )}{24 c^2}+\frac{5 d \sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right )}{8 c}-\frac{5 (a d+b c) \left (a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 \sqrt{a} \sqrt{c}}-\frac{(a+b x)^{5/2} (c+d x)^{5/2}}{3 x^3}-\frac{5 (a+b x)^{3/2} (c+d x)^{5/2} (a d+b c)}{12 c x^2}+\frac{5}{4} \sqrt{b} \sqrt{d} (a d+3 b c) (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right ) \]

[Out]

(5*d*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*c) + (5*
d*(9*b^2*c^2 + 14*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(24*c^2) - (
5*(3*b^2*c^2 + 12*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(24*c^2*x) -
(5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(5/2))/(12*c*x^2) - ((a + b*x)^(5/2)*(
c + d*x)^(5/2))/(3*x^3) - (5*(b*c + a*d)*(b^2*c^2 + 14*a*b*c*d + a^2*d^2)*ArcTan
h[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*Sqrt[a]*Sqrt[c]) + (5*Sqr
t[b]*Sqrt[d]*(3*b*c + a*d)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b
]*Sqrt[c + d*x])])/4

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Rubi [A]  time = 1.27053, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318 \[ -\frac{5 \sqrt{a+b x} (c+d x)^{5/2} \left (a^2 d^2+12 a b c d+3 b^2 c^2\right )}{24 c^2 x}+\frac{5 d \sqrt{a+b x} (c+d x)^{3/2} \left (a^2 d^2+14 a b c d+9 b^2 c^2\right )}{24 c^2}+\frac{5 d \sqrt{a+b x} \sqrt{c+d x} \left (a^2 d^2+10 a b c d+5 b^2 c^2\right )}{8 c}-\frac{5 (a d+b c) \left (a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 \sqrt{a} \sqrt{c}}-\frac{(a+b x)^{5/2} (c+d x)^{5/2}}{3 x^3}-\frac{5 (a+b x)^{3/2} (c+d x)^{5/2} (a d+b c)}{12 c x^2}+\frac{5}{4} \sqrt{b} \sqrt{d} (a d+3 b c) (3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]  Int[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^4,x]

[Out]

(5*d*(5*b^2*c^2 + 10*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*c) + (5*
d*(9*b^2*c^2 + 14*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(24*c^2) - (
5*(3*b^2*c^2 + 12*a*b*c*d + a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(24*c^2*x) -
(5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(5/2))/(12*c*x^2) - ((a + b*x)^(5/2)*(
c + d*x)^(5/2))/(3*x^3) - (5*(b*c + a*d)*(b^2*c^2 + 14*a*b*c*d + a^2*d^2)*ArcTan
h[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*Sqrt[a]*Sqrt[c]) + (5*Sqr
t[b]*Sqrt[d]*(3*b*c + a*d)*(b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b
]*Sqrt[c + d*x])])/4

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Rubi in Sympy [A]  time = 175.863, size = 330, normalized size = 0.97 \[ \frac{5 \sqrt{b} \sqrt{d} \left (a d + 3 b c\right ) \left (3 a d + b c\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{c + d x}}{\sqrt{d} \sqrt{a + b x}} \right )}}{4} - \frac{\left (a + b x\right )^{\frac{5}{2}} \left (c + d x\right )^{\frac{5}{2}}}{3 x^{3}} + \frac{5 d \sqrt{a + b x} \sqrt{c + d x} \left (a^{2} d^{2} + 10 a b c d + 5 b^{2} c^{2}\right )}{8 c} - \frac{5 \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{5}{2}} \left (a d + b c\right )}{12 c x^{2}} + \frac{5 d \sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}} \left (a^{2} d^{2} + 14 a b c d + 9 b^{2} c^{2}\right )}{24 c^{2}} - \frac{5 \sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}} \left (a^{2} d^{2} + 12 a b c d + 3 b^{2} c^{2}\right )}{24 c^{2} x} - \frac{5 \left (a d + b c\right ) \left (a^{2} d^{2} + 14 a b c d + b^{2} c^{2}\right ) \operatorname{atanh}{\left (\frac{\sqrt{c} \sqrt{a + b x}}{\sqrt{a} \sqrt{c + d x}} \right )}}{8 \sqrt{a} \sqrt{c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x+a)**(5/2)*(d*x+c)**(5/2)/x**4,x)

[Out]

5*sqrt(b)*sqrt(d)*(a*d + 3*b*c)*(3*a*d + b*c)*atanh(sqrt(b)*sqrt(c + d*x)/(sqrt(
d)*sqrt(a + b*x)))/4 - (a + b*x)**(5/2)*(c + d*x)**(5/2)/(3*x**3) + 5*d*sqrt(a +
b*x)*sqrt(c + d*x)*(a**2*d**2 + 10*a*b*c*d + 5*b**2*c**2)/(8*c) - 5*(a + b*x)**
(3/2)*(c + d*x)**(5/2)*(a*d + b*c)/(12*c*x**2) + 5*d*sqrt(a + b*x)*(c + d*x)**(3
/2)*(a**2*d**2 + 14*a*b*c*d + 9*b**2*c**2)/(24*c**2) - 5*sqrt(a + b*x)*(c + d*x)
**(5/2)*(a**2*d**2 + 12*a*b*c*d + 3*b**2*c**2)/(24*c**2*x) - 5*(a*d + b*c)*(a**2
*d**2 + 14*a*b*c*d + b**2*c**2)*atanh(sqrt(c)*sqrt(a + b*x)/(sqrt(a)*sqrt(c + d*
x)))/(8*sqrt(a)*sqrt(c))

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Mathematica [A]  time = 0.267033, size = 321, normalized size = 0.95 \[ \frac{1}{48} \left (-\frac{2 \sqrt{a+b x} \sqrt{c+d x} \left (a^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )+2 a b x \left (13 c^2+61 c d x-27 d^2 x^2\right )-3 b^2 x^2 \left (-11 c^2+18 c d x+4 d^2 x^2\right )\right )}{x^3}+30 \sqrt{b} \sqrt{d} \left (3 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )+\frac{15 \log (x) \left (a^3 d^3+15 a^2 b c d^2+15 a b^2 c^2 d+b^3 c^3\right )}{\sqrt{a} \sqrt{c}}-\frac{15 \left (a^3 d^3+15 a^2 b c d^2+15 a b^2 c^2 d+b^3 c^3\right ) \log \left (2 \sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x}+2 a c+a d x+b c x\right )}{\sqrt{a} \sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^4,x]

[Out]

((-2*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*b*x*(13*c^2 + 61*c*d*x - 27*d^2*x^2) - 3*b
^2*x^2*(-11*c^2 + 18*c*d*x + 4*d^2*x^2) + a^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/
x^3 + (15*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*Log[x])/(Sqrt[a]
*Sqrt[c]) - (15*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*Log[2*a*c
+ b*c*x + a*d*x + 2*Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]])/(Sqrt[a]*Sqrt[
c]) + 30*Sqrt[b]*Sqrt[d]*(3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*Log[b*c + a*d + 2*
b*d*x + 2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]])/48

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Maple [B]  time = 0.027, size = 848, normalized size = 2.5 \[ -{\frac{1}{48\,{x}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{a}^{3}{d}^{3}\sqrt{bd}+225\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{a}^{2}bc{d}^{2}\sqrt{bd}+225\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}a{b}^{2}{c}^{2}d\sqrt{bd}+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{3}{b}^{3}{c}^{3}\sqrt{bd}-90\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{3}{a}^{2}b{d}^{3}\sqrt{ac}-300\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{3}a{b}^{2}c{d}^{2}\sqrt{ac}-90\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{3}{b}^{3}{c}^{2}d\sqrt{ac}-24\,{x}^{4}{b}^{2}{d}^{2}\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-108\,{x}^{3}ab{d}^{2}\sqrt{bd}\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}-108\,{x}^{3}{b}^{2}cd\sqrt{bd}\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+66\,\sqrt{d{x}^{2}b+adx+bcx+ac}{d}^{2}\sqrt{bd}{a}^{2}\sqrt{ac}{x}^{2}+244\,\sqrt{d{x}^{2}b+adx+bcx+ac}db\sqrt{bd}a\sqrt{ac}{x}^{2}c+66\,{c}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}{b}^{2}\sqrt{bd}\sqrt{ac}{x}^{2}+52\,\sqrt{d{x}^{2}b+adx+bcx+ac}d\sqrt{bd}{a}^{2}\sqrt{ac}xc+52\,{c}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}b\sqrt{bd}a\sqrt{ac}x+16\,{c}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}{a}^{2}\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^4,x)

[Out]

-1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d
*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*a^3*d^3*(b*d)^(1/2)+225*ln((a*d*x+b*c*x+2*(a*c
)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*a^2*b*c*d^2*(b*d)^(1/2)+22
5*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^3*a*
b^2*c^2*d*(b*d)^(1/2)+15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)+2*a*c)/x)*x^3*b^3*c^3*(b*d)^(1/2)-90*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c
*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^3*a^2*b*d^3*(a*c)^(1/2)-300*ln
(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2)
)*x^3*a*b^2*c*d^2*(a*c)^(1/2)-90*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/
2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^3*b^3*c^2*d*(a*c)^(1/2)-24*x^4*b^2*d^2*(a
*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-108*x^3*a*b*d^2*(b*d)^(1/2
)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)-108*x^3*b^2*c*d*(b*d)^(1/2)*(a*c)^
(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*d^2*(b*
d)^(1/2)*a^2*(a*c)^(1/2)*x^2+244*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*d*b*(b*d)^(1/2)
*a*(a*c)^(1/2)*x^2*c+66*c^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^2*(b*d)^(1/2)*(a*c
)^(1/2)*x^2+52*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*d*(b*d)^(1/2)*a^2*(a*c)^(1/2)*x*c
+52*c^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b*(b*d)^(1/2)*a*(a*c)^(1/2)*x+16*c^2*(b*
d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*a^2*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)/x^3/(a*c)^(1/2)/(b*d)^(1/2)

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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 + a)^(5/2)*(d*x + c)^(5/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.92752, size = 1, normalized size = 0. \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^(5/2)*(d*x + c)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(30*(3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*sqrt(a*c)*sqrt(b*d)*x^3*log(8*b^2
*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqr
t(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 15*(b^3*c^3 + 15*a*b^2*c^2
*d + 15*a^2*b*c*d^2 + a^3*d^3)*x^3*log(-(4*(2*a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)*s
qrt(b*x + a)*sqrt(d*x + c) - (8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 +
8*(a*b*c^2 + a^2*c*d)*x)*sqrt(a*c))/x^2) + 4*(12*b^2*d^2*x^4 - 8*a^2*c^2 + 54*(b
^2*c*d + a*b*d^2)*x^3 - (33*b^2*c^2 + 122*a*b*c*d + 33*a^2*d^2)*x^2 - 26*(a*b*c^
2 + a^2*c*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c))/(sqrt(a*c)*x^3), 1/96*(60
*(3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*sqrt(a*c)*sqrt(-b*d)*x^3*arctan(1/2*(2*b*d
*x + b*c + a*d)/(sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c))) + 15*(b^3*c^3 + 15*a*b
^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*x^3*log(-(4*(2*a^2*c^2 + (a*b*c^2 + a^2*c*d
)*x)*sqrt(b*x + a)*sqrt(d*x + c) - (8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*
x^2 + 8*(a*b*c^2 + a^2*c*d)*x)*sqrt(a*c))/x^2) + 4*(12*b^2*d^2*x^4 - 8*a^2*c^2 +
54*(b^2*c*d + a*b*d^2)*x^3 - (33*b^2*c^2 + 122*a*b*c*d + 33*a^2*d^2)*x^2 - 26*(
a*b*c^2 + a^2*c*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c))/(sqrt(a*c)*x^3), 1/
48*(15*(3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*sqrt(-a*c)*sqrt(b*d)*x^3*log(8*b^2*d
^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(
b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 15*(b^3*c^3 + 15*a*b^2*c^2*d
+ 15*a^2*b*c*d^2 + a^3*d^3)*x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)/(
sqrt(b*x + a)*sqrt(d*x + c)*a*c)) + 2*(12*b^2*d^2*x^4 - 8*a^2*c^2 + 54*(b^2*c*d
+ a*b*d^2)*x^3 - (33*b^2*c^2 + 122*a*b*c*d + 33*a^2*d^2)*x^2 - 26*(a*b*c^2 + a^2
*c*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c))/(sqrt(-a*c)*x^3), 1/48*(30*(3*b
^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*sqrt(-a*c)*sqrt(-b*d)*x^3*arctan(1/2*(2*b*d*x +
b*c + a*d)/(sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c))) - 15*(b^3*c^3 + 15*a*b^2*c
^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*
c)/(sqrt(b*x + a)*sqrt(d*x + c)*a*c)) + 2*(12*b^2*d^2*x^4 - 8*a^2*c^2 + 54*(b^2*
c*d + a*b*d^2)*x^3 - (33*b^2*c^2 + 122*a*b*c*d + 33*a^2*d^2)*x^2 - 26*(a*b*c^2 +
a^2*c*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c))/(sqrt(-a*c)*x^3)]

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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+a)**(5/2)*(d*x+c)**(5/2)/x**4,x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.830312, size = 4, normalized size = 0.01 \[ \mathit{sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x + a)^(5/2)*(d*x + c)^(5/2)/x^4,x, algorithm="giac")

[Out]

sage0*x