3.1562 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=186 \[ -\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{2 e^4 (a+b x) (d+e x)^2}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^4 (a+b x)}+\frac{b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0856378, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ -\frac{3 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{2 e^4 (a+b x) (d+e x)^2}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^4 (a+b x)}+\frac{b^3 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0680947, size = 131, normalized size = 0.7 \[ -\frac{\sqrt{(a+b x)^2} \left (3 a^2 b e^2 (d+2 e x)+a^3 e^3-3 a b^2 d e (3 d+4 e x)+6 b^2 (d+e x)^2 (b d-a e) \log (d+e x)+b^3 \left (4 d^2 e x+5 d^3-4 d e^2 x^2-2 e^3 x^3\right )\right )}{2 e^4 (a+b x) (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(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 + 3*a^2*b*e^2*(d + 2*e*x) - 3*a*b^2*d*e*(3*d + 4*e*x) + b^3*(5*d^3 + 4*d^2*e*x -
4*d*e^2*x^2 - 2*e^3*x^3) + 6*b^2*(b*d - a*e)*(d + e*x)^2*Log[d + e*x]))/(2*e^4*(a + b*x)*(d + e*x)^2)

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Maple [A]  time = 0.202, size = 219, normalized size = 1.2 \begin{align*}{\frac{6\,\ln \left ( ex+d \right ){x}^{2}a{b}^{2}{e}^{3}-6\,\ln \left ( ex+d \right ){x}^{2}{b}^{3}d{e}^{2}+2\,{x}^{3}{b}^{3}{e}^{3}+12\,\ln \left ( ex+d \right ) xa{b}^{2}d{e}^{2}-12\,\ln \left ( ex+d \right ) x{b}^{3}{d}^{2}e+4\,{x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( ex+d \right ) a{b}^{2}{d}^{2}e-6\,\ln \left ( ex+d \right ){b}^{3}{d}^{3}-6\,x{a}^{2}b{e}^{3}+12\,xa{b}^{2}d{e}^{2}-4\,x{b}^{3}{d}^{2}e-{a}^{3}{e}^{3}-3\,d{e}^{2}{a}^{2}b+9\,a{b}^{2}{d}^{2}e-5\,{b}^{3}{d}^{3}}{2\, \left ( bx+a \right ) ^{3}{e}^{4} \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^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)*(6*ln(e*x+d)*x^2*a*b^2*e^3-6*ln(e*x+d)*x^2*b^3*d*e^2+2*x^3*b^3*e^3+12*ln(e*x+d)*x*a*b^2*
d*e^2-12*ln(e*x+d)*x*b^3*d^2*e+4*x^2*b^3*d*e^2+6*ln(e*x+d)*a*b^2*d^2*e-6*ln(e*x+d)*b^3*d^3-6*x*a^2*b*e^3+12*x*
a*b^2*d*e^2-4*x*b^3*d^2*e-a^3*e^3-3*d*e^2*a^2*b+9*a*b^2*d^2*e-5*b^3*d^3)/(b*x+a)^3/e^4/(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^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 = 1.50743, size = 375, normalized size = 2.02 \begin{align*} \frac{2 \, b^{3} e^{3} x^{3} + 4 \, b^{3} d e^{2} x^{2} - 5 \, b^{3} d^{3} + 9 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - a^{3} e^{3} - 2 \,{\left (2 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x - 6 \,{\left (b^{3} d^{3} - a b^{2} d^{2} e +{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (b^{3} d^{2} e - a b^{2} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.21325, size = 230, normalized size = 1.24 \begin{align*} b^{3} x e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) - 3 \,{\left (b^{3} d \mathrm{sgn}\left (b x + a\right ) - a b^{2} e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (5 \, b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 9 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) + a^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 6 \,{\left (b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) - 2 \, a b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) + a^{2} b e^{3} \mathrm{sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-4\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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