3.1729 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=284 \[ -\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x) (d+e x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{2 e^5 (a+b x) (d+e x)^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^3}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x) (-3 a B e-A b e+4 b B d)}{e^5 (a+b x)}+\frac{b^3 B x \sqrt{a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)} \]

[Out]

(b^3*B*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])/(3*e^5*(a + b*x)*(d + e*x)^3) + ((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]
)/(2*e^5*(a + b*x)*(d + e*x)^2) - (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])/(e
^5*(a + b*x)*(d + e*x)) - (b^2*(4*b*B*d - A*b*e - 3*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 = 0.211729, antiderivative size = 284, 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} \[ -\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x) (d+e x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{2 e^5 (a+b x) (d+e x)^2}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^3}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x) (-3 a B e-A b e+4 b B d)}{e^5 (a+b x)}+\frac{b^3 B x \sqrt{a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

(b^3*B*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])/(3*e^5*(a + b*x)*(d + e*x)^3) + ((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]
)/(2*e^5*(a + b*x)*(d + e*x)^2) - (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])/(e
^5*(a + b*x)*(d + e*x)) - (b^2*(4*b*B*d - A*b*e - 3*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)^4} \, 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)^4} \, 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^6 B}{e^4}-\frac{b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)^4}+\frac{b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^3}-\frac{3 b^4 (b d-a e) (-2 b B d+A b e+a B e)}{e^4 (d+e x)^2}+\frac{b^5 (-4 b B d+A b e+3 a B e)}{e^4 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{b^3 B 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}}{3 e^5 (a+b x) (d+e x)^3}+\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}}{2 e^5 (a+b x) (d+e x)^2}-\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}}{e^5 (a+b x) (d+e x)}-\frac{b^2 (4 b B d-A b e-3 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 = 0.164623, size = 251, normalized size = 0.88 \[ -\frac{\sqrt{(a+b x)^2} \left (3 a^2 b e^2 \left (A e (d+3 e x)+2 B \left (d^2+3 d e x+3 e^2 x^2\right )\right )+a^3 e^3 (2 A e+B (d+3 e x))+3 a b^2 e \left (2 A e \left (d^2+3 d e x+3 e^2 x^2\right )-B d \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 b^2 (d+e x)^3 \log (d+e x) (-3 a B e-A b e+4 b B d)+b^3 \left (2 B \left (9 d^2 e^2 x^2+27 d^3 e x+13 d^4-9 d e^3 x^3-3 e^4 x^4\right )-A d e \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )\right )}{6 e^5 (a+b x) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

-(Sqrt[(a + b*x)^2]*(a^3*e^3*(2*A*e + B*(d + 3*e*x)) + 3*a^2*b*e^2*(A*e*(d + 3*e*x) + 2*B*(d^2 + 3*d*e*x + 3*e
^2*x^2)) + 3*a*b^2*e*(2*A*e*(d^2 + 3*d*e*x + 3*e^2*x^2) - B*d*(11*d^2 + 27*d*e*x + 18*e^2*x^2)) + b^3*(-(A*d*e
*(11*d^2 + 27*d*e*x + 18*e^2*x^2)) + 2*B*(13*d^4 + 27*d^3*e*x + 9*d^2*e^2*x^2 - 9*d*e^3*x^3 - 3*e^4*x^4)) + 6*
b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^3*Log[d + e*x]))/(6*e^5*(a + b*x)*(d + e*x)^3)

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Maple [B]  time = 0.019, size = 512, normalized size = 1.8 \begin{align*}{\frac{54\,B{x}^{2}a{b}^{2}d{e}^{3}-54\,Bx{b}^{3}{d}^{3}e+27\,Ax{b}^{3}{d}^{2}{e}^{2}-9\,Ax{a}^{2}b{e}^{4}+6\,A\ln \left ( ex+d \right ){b}^{3}{d}^{3}e-18\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}-18\,B{x}^{2}{a}^{2}b{e}^{4}+18\,B{x}^{3}{b}^{3}d{e}^{3}-18\,A{x}^{2}a{b}^{2}{e}^{4}+18\,A{x}^{2}{b}^{3}d{e}^{3}-3\,Ad{e}^{3}{a}^{2}b-2\,A{a}^{3}{e}^{4}-26\,B{b}^{3}{d}^{4}+33\,Ba{b}^{2}{d}^{3}e-6\,B{a}^{2}b{d}^{2}{e}^{2}-6\,Aa{b}^{2}{d}^{2}{e}^{2}+54\,B\ln \left ( ex+d \right ){x}^{2}a{b}^{2}d{e}^{3}+54\,B\ln \left ( ex+d \right ) xa{b}^{2}{d}^{2}{e}^{2}-72\,B\ln \left ( ex+d \right ){x}^{2}{b}^{3}{d}^{2}{e}^{2}+18\,A\ln \left ( ex+d \right ){x}^{2}{b}^{3}d{e}^{3}+18\,A\ln \left ( ex+d \right ) x{b}^{3}{d}^{2}{e}^{2}-72\,B\ln \left ( ex+d \right ) x{b}^{3}{d}^{3}e-24\,B\ln \left ( ex+d \right ){x}^{3}{b}^{3}d{e}^{3}+18\,B\ln \left ( ex+d \right ){x}^{3}a{b}^{2}{e}^{4}-18\,Bx{a}^{2}bd{e}^{3}+81\,Bxa{b}^{2}{d}^{2}{e}^{2}-18\,Axa{b}^{2}d{e}^{3}+18\,B\ln \left ( ex+d \right ) a{b}^{2}{d}^{3}e+6\,B{x}^{4}{b}^{3}{e}^{4}-24\,B\ln \left ( ex+d \right ){b}^{3}{d}^{4}-3\,Bx{a}^{3}{e}^{4}-Bd{e}^{3}{a}^{3}+11\,A{b}^{3}{d}^{3}e+6\,A\ln \left ( ex+d \right ){x}^{3}{b}^{3}{e}^{4}}{6\, \left ( bx+a \right ) ^{3}{e}^{5} \left ( ex+d \right ) ^{3}} \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)^4,x)

[Out]

1/6*((b*x+a)^2)^(3/2)*(54*B*x^2*a*b^2*d*e^3-54*B*x*b^3*d^3*e+27*A*x*b^3*d^2*e^2-9*A*x*a^2*b*e^4+6*A*ln(e*x+d)*
b^3*d^3*e-18*B*x^2*b^3*d^2*e^2-18*B*x^2*a^2*b*e^4+18*B*x^3*b^3*d*e^3-18*A*x^2*a*b^2*e^4+18*A*x^2*b^3*d*e^3-3*A
*d*e^3*a^2*b-2*A*a^3*e^4-26*B*b^3*d^4+33*B*a*b^2*d^3*e-6*B*a^2*b*d^2*e^2-6*A*a*b^2*d^2*e^2+54*B*ln(e*x+d)*x^2*
a*b^2*d*e^3+54*B*ln(e*x+d)*x*a*b^2*d^2*e^2-72*B*ln(e*x+d)*x^2*b^3*d^2*e^2+18*A*ln(e*x+d)*x^2*b^3*d*e^3+18*A*ln
(e*x+d)*x*b^3*d^2*e^2-72*B*ln(e*x+d)*x*b^3*d^3*e-24*B*ln(e*x+d)*x^3*b^3*d*e^3+18*B*ln(e*x+d)*x^3*a*b^2*e^4-18*
B*x*a^2*b*d*e^3+81*B*x*a*b^2*d^2*e^2-18*A*x*a*b^2*d*e^3+18*B*ln(e*x+d)*a*b^2*d^3*e+6*B*x^4*b^3*e^4-24*B*ln(e*x
+d)*b^3*d^4-3*B*x*a^3*e^4-B*d*e^3*a^3+11*A*b^3*d^3*e+6*A*ln(e*x+d)*x^3*b^3*e^4)/(b*x+a)^3/e^5/(e*x+d)^3

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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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78806, size = 830, normalized size = 2.92 \begin{align*} \frac{6 \, B b^{3} e^{4} x^{4} + 18 \, B b^{3} d e^{3} x^{3} - 26 \, B b^{3} d^{4} - 2 \, A a^{3} e^{4} + 11 \,{\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} - 18 \,{\left (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} - 3 \,{\left (18 \, B b^{3} d^{3} e - 9 \,{\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 (4 \, B b^{3} d^{4} -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e +{\left (4 \, B b^{3} d e^{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} -{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3}\right )} x^{2} + 3 \,{\left (4 \, B b^{3} d^{3} e -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{6 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} 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)^4,x, algorithm="fricas")

[Out]

1/6*(6*B*b^3*e^4*x^4 + 18*B*b^3*d*e^3*x^3 - 26*B*b^3*d^4 - 2*A*a^3*e^4 + 11*(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 - 18*(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 - 3*(18*B*b^3*d^3*e - 9*(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*(4*B*b^3*d^4 - (3*B*a*b^2 + A*b^3)*d^3*e + (4*B*b^3*d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)
*x^3 + 3*(4*B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)*d*e^3)*x^2 + 3*(4*B*b^3*d^3*e - (3*B*a*b^2 + A*b^3)*d^2*e^2)*x
)*log(e*x + d))/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

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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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.15515, size = 555, normalized size = 1.95 \begin{align*} B b^{3} x e^{\left (-4\right )} \mathrm{sgn}\left (b x + a\right ) -{\left (4 \, B b^{3} d \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b^{2} e \mathrm{sgn}\left (b x + a\right ) - A b^{3} e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (26 \, B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) - 33 \, B a b^{2} d^{3} e \mathrm{sgn}\left (b x + a\right ) - 11 \, A b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 6 \, B a^{2} b d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, 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 ) + 2 \, A a^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 18 \,{\left (2 \, B b^{3} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e^{3} \mathrm{sgn}\left (b x + a\right ) - A b^{3} d e^{3} \mathrm{sgn}\left (b x + a\right ) + B a^{2} b e^{4} \mathrm{sgn}\left (b x + a\right ) + A a b^{2} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} x^{2} + 3 \,{\left (20 \, B b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) - 27 \, B a b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 9 \, 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 )}}{6 \,{\left (x e + d\right )}^{3}} \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)^4,x, algorithm="giac")

[Out]

B*b^3*x*e^(-4)*sgn(b*x + a) - (4*B*b^3*d*sgn(b*x + a) - 3*B*a*b^2*e*sgn(b*x + a) - A*b^3*e*sgn(b*x + a))*e^(-5
)*log(abs(x*e + d)) - 1/6*(26*B*b^3*d^4*sgn(b*x + a) - 33*B*a*b^2*d^3*e*sgn(b*x + a) - 11*A*b^3*d^3*e*sgn(b*x
+ a) + 6*B*a^2*b*d^2*e^2*sgn(b*x + a) + 6*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) + 2*A*a^3*e^4*sgn(b*x + a) + 18*(2*B*b^3*d^2*e^2*sgn(b*x + a) - 3*B*a*b^2*d*e^3*sgn(b*x + a
) - A*b^3*d*e^3*sgn(b*x + a) + B*a^2*b*e^4*sgn(b*x + a) + A*a*b^2*e^4*sgn(b*x + a))*x^2 + 3*(20*B*b^3*d^3*e*sg
n(b*x + a) - 27*B*a*b^2*d^2*e^2*sgn(b*x + a) - 9*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)^3