3.1207 \(\int \frac{(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=220 \[ -\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )+x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3-3 b^4 B e^3+2 b^3 B c d e^2\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{2 B e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

(-2*(d + e*x)^2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*
(b*x + c*x^2)^(3/2)) + (2*(b*c*d^2*(8*A*c^2*d + b^2*B*e - 4*b*c*(B*d + 2*A*e)) +
(16*A*c^4*d^3 + 2*b^3*B*c*d*e^2 - 3*b^4*B*e^3 - 8*b*c^3*d^2*(B*d + 3*A*e) + 2*b
^2*c^2*d*e*(3*B*d + 4*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) + (2*B*e^3*ArcTanh
[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(5/2)

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Rubi [A]  time = 0.538937, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ -\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )+x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3-3 b^4 B e^3+2 b^3 B c d e^2\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{2 B e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

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

[Out]

(-2*(d + e*x)^2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*
(b*x + c*x^2)^(3/2)) + (2*(b*c*d^2*(8*A*c^2*d + b^2*B*e - 4*b*c*(B*d + 2*A*e)) +
(16*A*c^4*d^3 + 2*b^3*B*c*d*e^2 - 3*b^4*B*e^3 - 8*b*c^3*d^2*(B*d + 3*A*e) + 2*b
^2*c^2*d*e*(3*B*d + 4*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) + (2*B*e^3*ArcTanh
[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(5/2)

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Rubi in Sympy [A]  time = 61.1182, size = 246, normalized size = 1.12 \[ \frac{2 B e^{3} \operatorname{atanh}{\left (\frac{\sqrt{c} x}{\sqrt{b x + c x^{2}}} \right )}}{c^{\frac{5}{2}}} - \frac{2 \left (d + e x\right )^{2} \left (A b c d + x \left (2 A c^{2} d + B b^{2} e - b c \left (A e + B d\right )\right )\right )}{3 b^{2} c \left (b x + c x^{2}\right )^{\frac{3}{2}}} + \frac{4 \left (\frac{b c d^{2} \left (- 8 A b c e + 8 A c^{2} d + B b^{2} e - 4 B b c d\right )}{2} + x \left (4 A b^{2} c^{2} d e^{2} - 12 A b c^{3} d^{2} e + 8 A c^{4} d^{3} - \frac{3 B b^{4} e^{3}}{2} + B b^{3} c d e^{2} + 3 B b^{2} c^{2} d^{2} e - 4 B b c^{3} d^{3}\right )\right )}{3 b^{4} c^{2} \sqrt{b x + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

2*B*e**3*atanh(sqrt(c)*x/sqrt(b*x + c*x**2))/c**(5/2) - 2*(d + e*x)**2*(A*b*c*d
+ x*(2*A*c**2*d + B*b**2*e - b*c*(A*e + B*d)))/(3*b**2*c*(b*x + c*x**2)**(3/2))
+ 4*(b*c*d**2*(-8*A*b*c*e + 8*A*c**2*d + B*b**2*e - 4*B*b*c*d)/2 + x*(4*A*b**2*c
**2*d*e**2 - 12*A*b*c**3*d**2*e + 8*A*c**4*d**3 - 3*B*b**4*e**3/2 + B*b**3*c*d*e
**2 + 3*B*b**2*c**2*d**2*e - 4*B*b*c**3*d**3))/(3*b**4*c**2*sqrt(b*x + c*x**2))

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Mathematica [A]  time = 0.970266, size = 199, normalized size = 0.9 \[ \frac{x^{5/2} \left (\frac{2 B e^3 (b+c x)^{5/2} \log \left (\sqrt{c} \sqrt{b+c x}+c \sqrt{x}\right )}{c^{5/2}}-\frac{2 (b+c x) \left (x^2 (b+c x) (c d-b e)^2 \left (b c (5 B d-A e)-8 A c^2 d+4 b^2 B e\right )+c^2 d^2 x (b+c x)^2 (9 A b e-8 A c d+3 b B d)+b x^2 (b B-A c) (c d-b e)^3+A b c^2 d^3 (b+c x)^2\right )}{3 b^4 c^2 x^{3/2}}\right )}{(x (b+c x))^{5/2}} \]

Antiderivative was successfully verified.

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

[Out]

(x^(5/2)*((-2*(b + c*x)*(b*(b*B - A*c)*(c*d - b*e)^3*x^2 + (c*d - b*e)^2*(-8*A*c
^2*d + 4*b^2*B*e + b*c*(5*B*d - A*e))*x^2*(b + c*x) + A*b*c^2*d^3*(b + c*x)^2 +
c^2*d^2*(3*b*B*d - 8*A*c*d + 9*A*b*e)*x*(b + c*x)^2))/(3*b^4*c^2*x^(3/2)) + (2*B
*e^3*(b + c*x)^(5/2)*Log[c*Sqrt[x] + Sqrt[c]*Sqrt[b + c*x]])/c^(5/2)))/(x*(b + c
*x))^(5/2)

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Maple [B]  time = 0.014, size = 680, normalized size = 3.1 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-16/3/b^3*c/(c*x^2+b*x)^(1/2)*x*B*d^3+2/b/(c*x^2+b*x)^(3/2)*x*A*d^2*e-2/c/(c*x^2
+b*x)^(3/2)*x*A*d*e^2-4/3*A*d^3/b^2/(c*x^2+b*x)^(3/2)*c*x-2/3*A*d^3/b/(c*x^2+b*x
)^(3/2)-8/3/b^2/(c*x^2+b*x)^(1/2)*B*d^3+1/3/c^2/(c*x^2+b*x)^(1/2)*A*e^3+B*e^3/c^
(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))-1/6*B*e^3*b/c^3/(c*x^2+b*x)^(1/2
)+1/6*B*e^3*b^2/c^3/(c*x^2+b*x)^(3/2)*x+1/2*B*e^3*b/c^2*x^2/(c*x^2+b*x)^(3/2)+2/
3/b/c/(c*x^2+b*x)^(1/2)*x*A*e^3-8/b^2/(c*x^2+b*x)^(1/2)*A*d^2*e-1/3*B*e^3*x^3/c/
(c*x^2+b*x)^(3/2)-7/3*B*e^3/c^2/(c*x^2+b*x)^(1/2)*x-x^2/c/(c*x^2+b*x)^(3/2)*A*e^
3+1/c^2/(c*x^2+b*x)^(1/2)*B*d*e^2+16/3*A*d^3*c/b^3/(c*x^2+b*x)^(1/2)+32/3*A*d^3*
c^2/b^4/(c*x^2+b*x)^(1/2)*x-3*x^2/c/(c*x^2+b*x)^(3/2)*B*d*e^2-1/3*b/c^2/(c*x^2+b
*x)^(3/2)*x*A*e^3-2/c/(c*x^2+b*x)^(3/2)*x*B*d^2*e+4/b^2/(c*x^2+b*x)^(1/2)*x*A*d*
e^2+4/b^2/(c*x^2+b*x)^(1/2)*x*B*d^2*e+2/b/c/(c*x^2+b*x)^(1/2)*A*d*e^2+2/b/c/(c*x
^2+b*x)^(1/2)*B*d^2*e+2/3/b/(c*x^2+b*x)^(3/2)*x*B*d^3-b/c^2/(c*x^2+b*x)^(3/2)*x*
B*d*e^2-16/b^3*c/(c*x^2+b*x)^(1/2)*x*A*d^2*e+2/b/c/(c*x^2+b*x)^(1/2)*x*B*d*e^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)*(e*x + d)^3/(c*x^2 + b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.308858, size = 1, normalized size = 0. \[ \left [\frac{3 \,{\left (B b^{4} c e^{3} x^{2} + B b^{5} e^{3} x\right )} \sqrt{c x^{2} + b x} \log \left ({\left (2 \, c x + b\right )} \sqrt{c} + 2 \, \sqrt{c x^{2} + b x} c\right ) - 2 \,{\left (A b^{3} c^{2} d^{3} +{\left (8 \,{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 6 \,{\left (B b^{2} c^{3} - 4 \, A b c^{4}\right )} d^{2} e - 3 \,{\left (B b^{3} c^{2} + 2 \, A b^{2} c^{3}\right )} d e^{2} +{\left (4 \, B b^{4} c - A b^{3} c^{2}\right )} e^{3}\right )} x^{3} - 3 \,{\left (3 \, A b^{3} c^{2} d e^{2} - B b^{5} e^{3} - 4 \,{\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} + 3 \,{\left (B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} d^{2} e\right )} x^{2} + 3 \,{\left (3 \, A b^{3} c^{2} d^{2} e +{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{3}\right )} x\right )} \sqrt{c}}{3 \,{\left (b^{4} c^{3} x^{2} + b^{5} c^{2} x\right )} \sqrt{c x^{2} + b x} \sqrt{c}}, \frac{2 \,{\left (3 \,{\left (B b^{4} c e^{3} x^{2} + B b^{5} e^{3} x\right )} \sqrt{c x^{2} + b x} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (A b^{3} c^{2} d^{3} +{\left (8 \,{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 6 \,{\left (B b^{2} c^{3} - 4 \, A b c^{4}\right )} d^{2} e - 3 \,{\left (B b^{3} c^{2} + 2 \, A b^{2} c^{3}\right )} d e^{2} +{\left (4 \, B b^{4} c - A b^{3} c^{2}\right )} e^{3}\right )} x^{3} - 3 \,{\left (3 \, A b^{3} c^{2} d e^{2} - B b^{5} e^{3} - 4 \,{\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} + 3 \,{\left (B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} d^{2} e\right )} x^{2} + 3 \,{\left (3 \, A b^{3} c^{2} d^{2} e +{\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d^{3}\right )} x\right )} \sqrt{-c}\right )}}{3 \,{\left (b^{4} c^{3} x^{2} + b^{5} c^{2} x\right )} \sqrt{c x^{2} + b x} \sqrt{-c}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/3*(3*(B*b^4*c*e^3*x^2 + B*b^5*e^3*x)*sqrt(c*x^2 + b*x)*log((2*c*x + b)*sqrt(c
) + 2*sqrt(c*x^2 + b*x)*c) - 2*(A*b^3*c^2*d^3 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 6*(
B*b^2*c^3 - 4*A*b*c^4)*d^2*e - 3*(B*b^3*c^2 + 2*A*b^2*c^3)*d*e^2 + (4*B*b^4*c -
A*b^3*c^2)*e^3)*x^3 - 3*(3*A*b^3*c^2*d*e^2 - B*b^5*e^3 - 4*(B*b^2*c^3 - 2*A*b*c^
4)*d^3 + 3*(B*b^3*c^2 - 4*A*b^2*c^3)*d^2*e)*x^2 + 3*(3*A*b^3*c^2*d^2*e + (B*b^3*
c^2 - 2*A*b^2*c^3)*d^3)*x)*sqrt(c))/((b^4*c^3*x^2 + b^5*c^2*x)*sqrt(c*x^2 + b*x)
*sqrt(c)), 2/3*(3*(B*b^4*c*e^3*x^2 + B*b^5*e^3*x)*sqrt(c*x^2 + b*x)*arctan(sqrt(
c*x^2 + b*x)*sqrt(-c)/(c*x)) - (A*b^3*c^2*d^3 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 6*(
B*b^2*c^3 - 4*A*b*c^4)*d^2*e - 3*(B*b^3*c^2 + 2*A*b^2*c^3)*d*e^2 + (4*B*b^4*c -
A*b^3*c^2)*e^3)*x^3 - 3*(3*A*b^3*c^2*d*e^2 - B*b^5*e^3 - 4*(B*b^2*c^3 - 2*A*b*c^
4)*d^3 + 3*(B*b^3*c^2 - 4*A*b^2*c^3)*d^2*e)*x^2 + 3*(3*A*b^3*c^2*d^2*e + (B*b^3*
c^2 - 2*A*b^2*c^3)*d^3)*x)*sqrt(-c))/((b^4*c^3*x^2 + b^5*c^2*x)*sqrt(c*x^2 + b*x
)*sqrt(-c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**3/(x*(b + c*x))**(5/2), x)

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GIAC/XCAS [A]  time = 0.309001, size = 390, normalized size = 1.77 \[ -\frac{B e^{3}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{5}{2}}} - \frac{2 \,{\left (\frac{A d^{3}}{b} +{\left (x{\left (\frac{{\left (8 \, B b c^{4} d^{3} - 16 \, A c^{5} d^{3} - 6 \, B b^{2} c^{3} d^{2} e + 24 \, A b c^{4} d^{2} e - 3 \, B b^{3} c^{2} d e^{2} - 6 \, A b^{2} c^{3} d e^{2} + 4 \, B b^{4} c e^{3} - A b^{3} c^{2} e^{3}\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (4 \, B b^{2} c^{3} d^{3} - 8 \, A b c^{4} d^{3} - 3 \, B b^{3} c^{2} d^{2} e + 12 \, A b^{2} c^{3} d^{2} e - 3 \, A b^{3} c^{2} d e^{2} + B b^{5} e^{3}\right )}}{b^{4} c^{2}}\right )} + \frac{3 \,{\left (B b^{3} c^{2} d^{3} - 2 \, A b^{2} c^{3} d^{3} + 3 \, A b^{3} c^{2} d^{2} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-B*e^3*ln(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) - 2/3*(A*
d^3/b + (x*((8*B*b*c^4*d^3 - 16*A*c^5*d^3 - 6*B*b^2*c^3*d^2*e + 24*A*b*c^4*d^2*e
- 3*B*b^3*c^2*d*e^2 - 6*A*b^2*c^3*d*e^2 + 4*B*b^4*c*e^3 - A*b^3*c^2*e^3)*x/(b^4
*c^2) + 3*(4*B*b^2*c^3*d^3 - 8*A*b*c^4*d^3 - 3*B*b^3*c^2*d^2*e + 12*A*b^2*c^3*d^
2*e - 3*A*b^3*c^2*d*e^2 + B*b^5*e^3)/(b^4*c^2)) + 3*(B*b^3*c^2*d^3 - 2*A*b^2*c^3
*d^3 + 3*A*b^3*c^2*d^2*e)/(b^4*c^2))*x)/(c*x^2 + b*x)^(3/2)