3.1209 \(\int \frac{(A+B x) (d+e x)}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=111 \[ \frac{2 (b+2 c x) \left (-4 b c (A e+B d)+8 A c^2 d+b^2 B e\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0979879, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {777, 613} \[ \frac{2 (b+2 c x) \left (-4 b c (A e+B d)+8 A c^2 d+b^2 B e\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}-\frac{\left (8 A c^2 d+b^2 B e-4 b c (B d+A e)\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2 c}\\ &=-\frac{2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (8 A c^2 d+b^2 B e-4 b c (B d+A e)\right ) (b+2 c x)}{3 b^4 c \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0526959, size = 107, normalized size = 0.96 \[ -\frac{2 \left (A \left (-6 b^2 c x (d-2 e x)+b^3 (d+3 e x)+8 b c^2 x^2 (e x-3 d)-16 c^3 d x^3\right )+b B x \left (3 b^2 (d-e x)-2 b c x (e x-6 d)+8 c^2 d x^2\right )\right )}{3 b^4 (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B*x*(8*c^2*d*x^2 + 3*b^2*(d - e*x) - 2*b*c*x*(-6*d + e*x)) + A*(-16*c^3*d*x^3 - 6*b^2*c*x*(d - 2*e*x) +
8*b*c^2*x^2*(-3*d + e*x) + b^3*(d + 3*e*x))))/(3*b^4*(x*(b + c*x))^(3/2))

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Maple [A]  time = 0.004, size = 141, normalized size = 1.3 \begin{align*} -{\frac{2\,x \left ( cx+b \right ) \left ( 8\,Ab{c}^{2}e{x}^{3}-16\,A{c}^{3}d{x}^{3}-2\,B{b}^{2}ce{x}^{3}+8\,Bb{c}^{2}d{x}^{3}+12\,A{b}^{2}ce{x}^{2}-24\,Ab{c}^{2}d{x}^{2}-3\,B{b}^{3}e{x}^{2}+12\,B{b}^{2}cd{x}^{2}+3\,A{b}^{3}ex-6\,A{b}^{2}cdx+3\,B{b}^{3}dx+Ad{b}^{3} \right ) }{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.996861, size = 285, normalized size = 2.57 \begin{align*} -\frac{4 \, A c d x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{2}} + \frac{32 \, A c^{2} d x}{3 \, \sqrt{c x^{2} + b x} b^{4}} + \frac{4 \, B e x}{3 \, \sqrt{c x^{2} + b x} b^{2}} - \frac{2 \, B e x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c} - \frac{2 \, A d}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} + \frac{16 \, A c d}{3 \, \sqrt{c x^{2} + b x} b^{3}} + \frac{2 \, B e}{3 \, \sqrt{c x^{2} + b x} b c} + \frac{2 \,{\left (B d + A e\right )} x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} - \frac{16 \,{\left (B d + A e\right )} c x}{3 \, \sqrt{c x^{2} + b x} b^{3}} - \frac{8 \,{\left (B d + A e\right )}}{3 \, \sqrt{c x^{2} + b x} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

-4/3*A*c*d*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*A*c^2*d*x/(sqrt(c*x^2 + b*x)*b^4) + 4/3*B*e*x/(sqrt(c*x^2 + b*x)
*b^2) - 2/3*B*e*x/((c*x^2 + b*x)^(3/2)*c) - 2/3*A*d/((c*x^2 + b*x)^(3/2)*b) + 16/3*A*c*d/(sqrt(c*x^2 + b*x)*b^
3) + 2/3*B*e/(sqrt(c*x^2 + b*x)*b*c) + 2/3*(B*d + A*e)*x/((c*x^2 + b*x)^(3/2)*b) - 16/3*(B*d + A*e)*c*x/(sqrt(
c*x^2 + b*x)*b^3) - 8/3*(B*d + A*e)/(sqrt(c*x^2 + b*x)*b^2)

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Fricas [A]  time = 1.71841, size = 313, normalized size = 2.82 \begin{align*} -\frac{2 \,{\left (A b^{3} d + 2 \,{\left (4 \,{\left (B b c^{2} - 2 \, A c^{3}\right )} d -{\left (B b^{2} c - 4 \, A b c^{2}\right )} e\right )} x^{3} + 3 \,{\left (4 \,{\left (B b^{2} c - 2 \, A b c^{2}\right )} d -{\left (B b^{3} - 4 \, A b^{2} c\right )} e\right )} x^{2} + 3 \,{\left (A b^{3} e +{\left (B b^{3} - 2 \, A b^{2} c\right )} d\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(A*b^3*d + 2*(4*(B*b*c^2 - 2*A*c^3)*d - (B*b^2*c - 4*A*b*c^2)*e)*x^3 + 3*(4*(B*b^2*c - 2*A*b*c^2)*d - (B*
b^3 - 4*A*b^2*c)*e)*x^2 + 3*(A*b^3*e + (B*b^3 - 2*A*b^2*c)*d)*x)*sqrt(c*x^2 + b*x)/(b^4*c^2*x^4 + 2*b^5*c*x^3
+ b^6*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22635, size = 194, normalized size = 1.75 \begin{align*} -\frac{{\left (x{\left (\frac{2 \,{\left (4 \, B b c^{2} d - 8 \, A c^{3} d - B b^{2} c e + 4 \, A b c^{2} e\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (4 \, B b^{2} c d - 8 \, A b c^{2} d - B b^{3} e + 4 \, A b^{2} c e\right )}}{b^{4} c^{2}}\right )} + \frac{3 \,{\left (B b^{3} d - 2 \, A b^{2} c d + A b^{3} e\right )}}{b^{4} c^{2}}\right )} x + \frac{A d}{b c^{2}}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((x*(2*(4*B*b*c^2*d - 8*A*c^3*d - B*b^2*c*e + 4*A*b*c^2*e)*x/(b^4*c^2) + 3*(4*B*b^2*c*d - 8*A*b*c^2*d - B
*b^3*e + 4*A*b^2*c*e)/(b^4*c^2)) + 3*(B*b^3*d - 2*A*b^2*c*d + A*b^3*e)/(b^4*c^2))*x + A*d/(b*c^2))/(c*x^2 + b*
x)^(3/2)