3.1728 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=280 \[ -\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+3 b B d)}{e^4 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) (d+e x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{2 e^5 (a+b x) (d+e x)^2}+\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac{b^3 B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.234223, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ -\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+3 b B d)}{e^4 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) (d+e x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{2 e^5 (a+b x) (d+e x)^2}+\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac{b^3 B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3 (A+B x)}{(d+e x)^3} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{b^5 (-3 b B d+A b e+3 a B e)}{e^4}+\frac{b^6 B x}{e^3}-\frac{b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)^3}+\frac{b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^2}-\frac{3 b^4 (b d-a e) (-2 b B d+A b e+a B e)}{e^4 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{b^2 (3 b B d-A b e-3 a B e) x \sqrt{a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac{b^3 B x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)}-\frac{(b d-a e)^3 (B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac{(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac{3 b (b d-a e) (2 b B d-A b e-a B e) \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.169924, size = 256, normalized size = 0.91 \[ \frac{\sqrt{(a+b x)^2} \left (-3 a^2 b e^2 (A e (d+2 e x)-B d (3 d+4 e x))-a^3 e^3 (A e+B (d+2 e x))+3 a b^2 e \left (A d e (3 d+4 e x)+B \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )\right )+6 b (d+e x)^2 (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)+b^3 \left (A e \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )+B \left (-11 d^2 e^2 x^2+2 d^3 e x+7 d^4-4 d e^3 x^3+e^4 x^4\right )\right )\right )}{2 e^5 (a+b x) (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-(a^3*e^3*(A*e + B*(d + 2*e*x))) - 3*a^2*b*e^2*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) + 3*a
*b^2*e*(A*d*e*(3*d + 4*e*x) + B*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + b^3*(A*e*(-5*d^3 - 4*d^2*e*x
+ 4*d*e^2*x^2 + 2*e^3*x^3) + B*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4)) + 6*b*(b*d - a*e
)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2*Log[d + e*x]))/(2*e^5*(a + b*x)*(d + e*x)^2)

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Maple [B]  time = 0.018, size = 566, normalized size = 2. \begin{align*}{\frac{12\,B{x}^{2}a{b}^{2}d{e}^{3}+2\,Bx{b}^{3}{d}^{3}e-4\,Ax{b}^{3}{d}^{2}{e}^{2}-6\,Ax{a}^{2}b{e}^{4}-6\,A\ln \left ( ex+d \right ){b}^{3}{d}^{3}e-11\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}-4\,B{x}^{3}{b}^{3}d{e}^{3}+4\,A{x}^{2}{b}^{3}d{e}^{3}+6\,B{x}^{3}a{b}^{2}{e}^{4}-3\,Ad{e}^{3}{a}^{2}b-A{a}^{3}{e}^{4}+7\,B{b}^{3}{d}^{4}-15\,Ba{b}^{2}{d}^{3}e+9\,B{a}^{2}b{d}^{2}{e}^{2}+9\,Aa{b}^{2}{d}^{2}{e}^{2}-18\,B\ln \left ( ex+d \right ){x}^{2}a{b}^{2}d{e}^{3}-36\,B\ln \left ( ex+d \right ) xa{b}^{2}{d}^{2}{e}^{2}+12\,B\ln \left ( ex+d \right ) x{a}^{2}bd{e}^{3}+6\,A\ln \left ( ex+d \right ){x}^{2}a{b}^{2}{e}^{4}+12\,B\ln \left ( ex+d \right ){x}^{2}{b}^{3}{d}^{2}{e}^{2}+6\,B\ln \left ( ex+d \right ){x}^{2}{a}^{2}b{e}^{4}-6\,A\ln \left ( ex+d \right ){x}^{2}{b}^{3}d{e}^{3}-12\,A\ln \left ( ex+d \right ) x{b}^{3}{d}^{2}{e}^{2}+24\,B\ln \left ( ex+d \right ) x{b}^{3}{d}^{3}e+12\,Bx{a}^{2}bd{e}^{3}-12\,Bxa{b}^{2}{d}^{2}{e}^{2}+6\,A\ln \left ( ex+d \right ) a{b}^{2}{d}^{2}{e}^{2}+12\,Axa{b}^{2}d{e}^{3}+6\,B\ln \left ( ex+d \right ){a}^{2}b{d}^{2}{e}^{2}-18\,B\ln \left ( ex+d \right ) a{b}^{2}{d}^{3}e+B{x}^{4}{b}^{3}{e}^{4}+2\,A{x}^{3}{b}^{3}{e}^{4}+12\,B\ln \left ( ex+d \right ){b}^{3}{d}^{4}-2\,Bx{a}^{3}{e}^{4}-Bd{e}^{3}{a}^{3}-5\,A{b}^{3}{d}^{3}e+12\,A\ln \left ( ex+d \right ) xa{b}^{2}d{e}^{3}}{2\, \left ( bx+a \right ) ^{3}{e}^{5} \left ( ex+d \right ) ^{2}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x)

[Out]

1/2*((b*x+a)^2)^(3/2)*(12*B*x^2*a*b^2*d*e^3+2*B*x*b^3*d^3*e-4*A*x*b^3*d^2*e^2-6*A*x*a^2*b*e^4-6*A*ln(e*x+d)*b^
3*d^3*e-11*B*x^2*b^3*d^2*e^2-4*B*x^3*b^3*d*e^3+4*A*x^2*b^3*d*e^3+6*B*x^3*a*b^2*e^4-3*A*d*e^3*a^2*b-A*a^3*e^4+7
*B*b^3*d^4-15*B*a*b^2*d^3*e+9*B*a^2*b*d^2*e^2+9*A*a*b^2*d^2*e^2-18*B*ln(e*x+d)*x^2*a*b^2*d*e^3-36*B*ln(e*x+d)*
x*a*b^2*d^2*e^2+12*B*ln(e*x+d)*x*a^2*b*d*e^3+6*A*ln(e*x+d)*x^2*a*b^2*e^4+12*B*ln(e*x+d)*x^2*b^3*d^2*e^2+6*B*ln
(e*x+d)*x^2*a^2*b*e^4-6*A*ln(e*x+d)*x^2*b^3*d*e^3-12*A*ln(e*x+d)*x*b^3*d^2*e^2+24*B*ln(e*x+d)*x*b^3*d^3*e+12*B
*x*a^2*b*d*e^3-12*B*x*a*b^2*d^2*e^2+6*A*ln(e*x+d)*a*b^2*d^2*e^2+12*A*x*a*b^2*d*e^3+6*B*ln(e*x+d)*a^2*b*d^2*e^2
-18*B*ln(e*x+d)*a*b^2*d^3*e+B*x^4*b^3*e^4+2*A*x^3*b^3*e^4+12*B*ln(e*x+d)*b^3*d^4-2*B*x*a^3*e^4-B*d*e^3*a^3-5*A
*b^3*d^3*e+12*A*ln(e*x+d)*x*a*b^2*d*e^3)/(b*x+a)^3/e^5/(e*x+d)^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9426, size = 849, normalized size = 3.03 \begin{align*} \frac{B b^{3} e^{4} x^{4} + 7 \, B b^{3} d^{4} - A a^{3} e^{4} - 5 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 9 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} -{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 2 \,{\left (2 \, B b^{3} d e^{3} -{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} -{\left (11 \, B b^{3} d^{2} e^{2} - 4 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3}\right )} x^{2} + 2 \,{\left (B b^{3} d^{3} e - 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} -{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 6 \,{\left (2 \, B b^{3} d^{4} -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e +{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} +{\left (2 \, B b^{3} d^{2} e^{2} -{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} +{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 2 \,{\left (2 \, B b^{3} d^{3} e -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} +{\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(B*b^3*e^4*x^4 + 7*B*b^3*d^4 - A*a^3*e^4 - 5*(3*B*a*b^2 + A*b^3)*d^3*e + 9*(B*a^2*b + A*a*b^2)*d^2*e^2 - (
B*a^3 + 3*A*a^2*b)*d*e^3 - 2*(2*B*b^3*d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)*x^3 - (11*B*b^3*d^2*e^2 - 4*(3*B*a*b^2
+ A*b^3)*d*e^3)*x^2 + 2*(B*b^3*d^3*e - 2*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 6*(B*a^2*b + A*a*b^2)*d*e^3 - (B*a^3 +
3*A*a^2*b)*e^4)*x + 6*(2*B*b^3*d^4 - (3*B*a*b^2 + A*b^3)*d^3*e + (B*a^2*b + A*a*b^2)*d^2*e^2 + (2*B*b^3*d^2*e^
2 - (3*B*a*b^2 + A*b^3)*d*e^3 + (B*a^2*b + A*a*b^2)*e^4)*x^2 + 2*(2*B*b^3*d^3*e - (3*B*a*b^2 + A*b^3)*d^2*e^2
+ (B*a^2*b + A*a*b^2)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**3,x)

[Out]

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

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Giac [A]  time = 1.15823, size = 563, normalized size = 2.01 \begin{align*} 3 \,{\left (2 \, B b^{3} d^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e \mathrm{sgn}\left (b x + a\right ) - A b^{3} d e \mathrm{sgn}\left (b x + a\right ) + B a^{2} b e^{2} \mathrm{sgn}\left (b x + a\right ) + A a b^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (B b^{3} x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) - 6 \, B b^{3} d x e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, B a b^{2} x e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \, A b^{3} x e^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} + \frac{{\left (7 \, B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) - 15 \, B a b^{2} d^{3} e \mathrm{sgn}\left (b x + a\right ) - 5 \, A b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 9 \, B a^{2} b d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 9 \, A a b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - B a^{3} d e^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} b d e^{3} \mathrm{sgn}\left (b x + a\right ) - A a^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (4 \, B b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \, A a b^{2} d e^{3} \mathrm{sgn}\left (b x + a\right ) - B a^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} b e^{4} \mathrm{sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-5\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

3*(2*B*b^3*d^2*sgn(b*x + a) - 3*B*a*b^2*d*e*sgn(b*x + a) - A*b^3*d*e*sgn(b*x + a) + B*a^2*b*e^2*sgn(b*x + a) +
A*a*b^2*e^2*sgn(b*x + a))*e^(-5)*log(abs(x*e + d)) + 1/2*(B*b^3*x^2*e^3*sgn(b*x + a) - 6*B*b^3*d*x*e^2*sgn(b*
x + a) + 6*B*a*b^2*x*e^3*sgn(b*x + a) + 2*A*b^3*x*e^3*sgn(b*x + a))*e^(-6) + 1/2*(7*B*b^3*d^4*sgn(b*x + a) - 1
5*B*a*b^2*d^3*e*sgn(b*x + a) - 5*A*b^3*d^3*e*sgn(b*x + a) + 9*B*a^2*b*d^2*e^2*sgn(b*x + a) + 9*A*a*b^2*d^2*e^2
*sgn(b*x + a) - B*a^3*d*e^3*sgn(b*x + a) - 3*A*a^2*b*d*e^3*sgn(b*x + a) - A*a^3*e^4*sgn(b*x + a) + 2*(4*B*b^3*
d^3*e*sgn(b*x + a) - 9*B*a*b^2*d^2*e^2*sgn(b*x + a) - 3*A*b^3*d^2*e^2*sgn(b*x + a) + 6*B*a^2*b*d*e^3*sgn(b*x +
a) + 6*A*a*b^2*d*e^3*sgn(b*x + a) - B*a^3*e^4*sgn(b*x + a) - 3*A*a^2*b*e^4*sgn(b*x + a))*x)*e^(-5)/(x*e + d)^
2