3.1727 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=285 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) (d+e x)}+\frac{3 b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^4 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}+\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.266737, antiderivative size = 285, 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} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) (d+e x)}+\frac{3 b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^4 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (-3 a B e-A b e+4 b B d)}{2 e^5 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}+\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.156341, size = 262, normalized size = 0.92 \[ \frac{\sqrt{(a+b x)^2} \left (18 a^2 b e^2 \left (A d e+B \left (-d^2+d e x+e^2 x^2\right )\right )+6 a^3 e^3 (B d-A e)+9 a b^2 e \left (2 A e \left (-d^2+d e x+e^2 x^2\right )+B \left (-4 d^2 e x+2 d^3-3 d e^2 x^2+e^3 x^3\right )\right )-6 (d+e x) (b d-a e)^2 \log (d+e x) (-a B e-3 A b e+4 b B d)+b^3 \left (3 A e \left (-4 d^2 e x+2 d^3-3 d e^2 x^2+e^3 x^3\right )+2 B \left (6 d^2 e^2 x^2+9 d^3 e x-3 d^4-2 d e^3 x^3+e^4 x^4\right )\right )\right )}{6 e^5 (a+b x) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(6*a^3*e^3*(B*d - A*e) + 18*a^2*b*e^2*(A*d*e + B*(-d^2 + d*e*x + e^2*x^2)) + 9*a*b^2*e*(2*A
*e*(-d^2 + d*e*x + e^2*x^2) + B*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3)) + b^3*(3*A*e*(2*d^3 - 4*d^2*e*x -
3*d*e^2*x^2 + e^3*x^3) + 2*B*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4)) - 6*(b*d - a*e)^2*
(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x)*Log[d + e*x]))/(6*e^5*(a + b*x)*(d + e*x))

________________________________________________________________________________________

Maple [B]  time = 0.019, size = 540, normalized size = 1.9 \begin{align*}{\frac{-27\,B{x}^{2}a{b}^{2}d{e}^{3}+6\,B\ln \left ( ex+d \right ){a}^{3}d{e}^{3}+18\,Bx{b}^{3}{d}^{3}e-12\,Ax{b}^{3}{d}^{2}{e}^{2}+18\,A\ln \left ( ex+d \right ){b}^{3}{d}^{3}e+12\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}+18\,B{x}^{2}{a}^{2}b{e}^{4}-4\,B{x}^{3}{b}^{3}d{e}^{3}+18\,A{x}^{2}a{b}^{2}{e}^{4}-9\,A{x}^{2}{b}^{3}d{e}^{3}+9\,B{x}^{3}a{b}^{2}{e}^{4}+18\,Ad{e}^{3}{a}^{2}b-6\,A{a}^{3}{e}^{4}-6\,B{b}^{3}{d}^{4}+18\,Ba{b}^{2}{d}^{3}e-18\,B{a}^{2}b{d}^{2}{e}^{2}-18\,Aa{b}^{2}{d}^{2}{e}^{2}+54\,B\ln \left ( ex+d \right ) xa{b}^{2}{d}^{2}{e}^{2}-36\,B\ln \left ( ex+d \right ) x{a}^{2}bd{e}^{3}+18\,A\ln \left ( ex+d \right ) x{a}^{2}b{e}^{4}+18\,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+18\,Bx{a}^{2}bd{e}^{3}-36\,Bxa{b}^{2}{d}^{2}{e}^{2}+18\,A\ln \left ( ex+d \right ){a}^{2}bd{e}^{3}-36\,A\ln \left ( ex+d \right ) a{b}^{2}{d}^{2}{e}^{2}+18\,Axa{b}^{2}d{e}^{3}-36\,B\ln \left ( ex+d \right ){a}^{2}b{d}^{2}{e}^{2}+54\,B\ln \left ( ex+d \right ) a{b}^{2}{d}^{3}e+2\,B{x}^{4}{b}^{3}{e}^{4}+3\,A{x}^{3}{b}^{3}{e}^{4}-24\,B\ln \left ( ex+d \right ){b}^{3}{d}^{4}+6\,Bd{e}^{3}{a}^{3}+6\,A{b}^{3}{d}^{3}e+6\,B\ln \left ( ex+d \right ) x{a}^{3}{e}^{4}-36\,A\ln \left ( ex+d \right ) xa{b}^{2}d{e}^{3}}{6\, \left ( bx+a \right ) ^{3}{e}^{5} \left ( ex+d \right ) } \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)^2,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(-27*B*x^2*a*b^2*d*e^3+6*B*ln(e*x+d)*a^3*d*e^3+18*B*x*b^3*d^3*e-12*A*x*b^3*d^2*e^2+18*A*
ln(e*x+d)*b^3*d^3*e+12*B*x^2*b^3*d^2*e^2+18*B*x^2*a^2*b*e^4-4*B*x^3*b^3*d*e^3+18*A*x^2*a*b^2*e^4-9*A*x^2*b^3*d
*e^3+9*B*x^3*a*b^2*e^4+18*A*d*e^3*a^2*b-6*A*a^3*e^4-6*B*b^3*d^4+18*B*a*b^2*d^3*e-18*B*a^2*b*d^2*e^2-18*A*a*b^2
*d^2*e^2+54*B*ln(e*x+d)*x*a*b^2*d^2*e^2-36*B*ln(e*x+d)*x*a^2*b*d*e^3+18*A*ln(e*x+d)*x*a^2*b*e^4+18*A*ln(e*x+d)
*x*b^3*d^2*e^2-24*B*ln(e*x+d)*x*b^3*d^3*e+18*B*x*a^2*b*d*e^3-36*B*x*a*b^2*d^2*e^2+18*A*ln(e*x+d)*a^2*b*d*e^3-3
6*A*ln(e*x+d)*a*b^2*d^2*e^2+18*A*x*a*b^2*d*e^3-36*B*ln(e*x+d)*a^2*b*d^2*e^2+54*B*ln(e*x+d)*a*b^2*d^3*e+2*B*x^4
*b^3*e^4+3*A*x^3*b^3*e^4-24*B*ln(e*x+d)*b^3*d^4+6*B*d*e^3*a^3+6*A*b^3*d^3*e+6*B*ln(e*x+d)*x*a^3*e^4-36*A*ln(e*
x+d)*x*a*b^2*d*e^3)/(b*x+a)^3/e^5/(e*x+d)

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.57629, size = 819, normalized size = 2.87 \begin{align*} \frac{2 \, B b^{3} e^{4} x^{4} - 6 \, B b^{3} d^{4} - 6 \, A a^{3} e^{4} + 6 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 18 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 6 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} -{\left (4 \, B b^{3} d e^{3} - 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \,{\left (4 \, B b^{3} d^{2} e^{2} - 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \,{\left (3 \, B b^{3} d^{3} e - 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x - 6 \,{\left (4 \, B b^{3} d^{4} - 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 6 \,{\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} +{\left (4 \, B b^{3} d^{3} e - 3 \,{\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\right )} \log \left (e x + d\right )}{6 \,{\left (e^{6} x + d 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)^2,x, algorithm="fricas")

[Out]

1/6*(2*B*b^3*e^4*x^4 - 6*B*b^3*d^4 - 6*A*a^3*e^4 + 6*(3*B*a*b^2 + A*b^3)*d^3*e - 18*(B*a^2*b + A*a*b^2)*d^2*e^
2 + 6*(B*a^3 + 3*A*a^2*b)*d*e^3 - (4*B*b^3*d*e^3 - 3*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(4*B*b^3*d^2*e^2 - 3*(3*
B*a*b^2 + A*b^3)*d*e^3 + 6*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 6*(3*B*b^3*d^3*e - 2*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 3
*(B*a^2*b + A*a*b^2)*d*e^3)*x - 6*(4*B*b^3*d^4 - 3*(3*B*a*b^2 + A*b^3)*d^3*e + 6*(B*a^2*b + A*a*b^2)*d^2*e^2 -
(B*a^3 + 3*A*a^2*b)*d*e^3 + (4*B*b^3*d^3*e - 3*(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)*log(e*x + d))/(e^6*x + d*e^5)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.17443, size = 574, normalized size = 2.01 \begin{align*} -{\left (4 \, B b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, A a b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - B a^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} b e^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{6} \,{\left (2 \, B b^{3} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) - 6 \, B b^{3} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, B b^{3} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 9 \, B a b^{2} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 3 \, A b^{3} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) - 36 \, B a b^{2} d x e^{3} \mathrm{sgn}\left (b x + a\right ) - 12 \, A b^{3} d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, B a^{2} b x e^{4} \mathrm{sgn}\left (b x + a\right ) + 18 \, A a b^{2} x e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} - \frac{{\left (B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b^{2} d^{3} e \mathrm{sgn}\left (b x + a\right ) - A b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 3 \, B a^{2} b d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, 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 )\right )} e^{\left (-5\right )}}{x e + d} \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)^2,x, algorithm="giac")

[Out]

-(4*B*b^3*d^3*sgn(b*x + a) - 9*B*a*b^2*d^2*e*sgn(b*x + a) - 3*A*b^3*d^2*e*sgn(b*x + a) + 6*B*a^2*b*d*e^2*sgn(b
*x + a) + 6*A*a*b^2*d*e^2*sgn(b*x + a) - B*a^3*e^3*sgn(b*x + a) - 3*A*a^2*b*e^3*sgn(b*x + a))*e^(-5)*log(abs(x
*e + d)) + 1/6*(2*B*b^3*x^3*e^4*sgn(b*x + a) - 6*B*b^3*d*x^2*e^3*sgn(b*x + a) + 18*B*b^3*d^2*x*e^2*sgn(b*x + a
) + 9*B*a*b^2*x^2*e^4*sgn(b*x + a) + 3*A*b^3*x^2*e^4*sgn(b*x + a) - 36*B*a*b^2*d*x*e^3*sgn(b*x + a) - 12*A*b^3
*d*x*e^3*sgn(b*x + a) + 18*B*a^2*b*x*e^4*sgn(b*x + a) + 18*A*a*b^2*x*e^4*sgn(b*x + a))*e^(-6) - (B*b^3*d^4*sgn
(b*x + a) - 3*B*a*b^2*d^3*e*sgn(b*x + a) - A*b^3*d^3*e*sgn(b*x + a) + 3*B*a^2*b*d^2*e^2*sgn(b*x + a) + 3*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))*e^
(-5)/(x*e + d)