3.1977 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=239 \[ \frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (a+b x)}-\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}{e^5 (a+b x)}+\frac{6 b^2 x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x) (d+e x)}-\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^5 (a+b x)} \]

[Out]

(6*b^2*(b*d - a*e)^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x)) - ((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(e^5*(a + b*x)*(d + e*x)) - (2*b^3*(b*d - a*e)*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*
x)) + (b^4*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)) - (4*b*(b*d - a*e)^3*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 = 0.16326, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3}{3 e^5 (a+b x)}-\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}{e^5 (a+b x)}+\frac{6 b^2 x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x) (d+e x)}-\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^5 (a+b x)} \]

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]

(6*b^2*(b*d - a*e)^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x)) - ((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(e^5*(a + b*x)*(d + e*x)) - (2*b^3*(b*d - a*e)*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*
x)) + (b^4*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)) - (4*b*(b*d - a*e)^3*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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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{(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{(a+b x) \left (a b+b^2 x\right )^3}{(d+e x)^2} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{(a+b x)^4}{(d+e x)^2} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{6 b^2 (b d-a e)^2}{e^4}+\frac{(-b d+a e)^4}{e^4 (d+e x)^2}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)}-\frac{4 b^3 (b d-a e) (d+e x)}{e^4}+\frac{b^4 (d+e x)^2}{e^4}\right ) \, dx}{a b+b^2 x}\\ &=\frac{6 b^2 (b d-a e)^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}-\frac{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}-\frac{2 b^3 (b d-a e) (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac{b^4 (d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}-\frac{4 b (b d-a e)^3 \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 = 0.125875, size = 183, normalized size = 0.77 \[ \frac{\sqrt{(a+b x)^2} \left (18 a^2 b^2 e^2 \left (-d^2+d e x+e^2 x^2\right )+12 a^3 b d e^3-3 a^4 e^4+6 a b^3 e \left (-4 d^2 e x+2 d^3-3 d e^2 x^2+e^3 x^3\right )-12 b (d+e x) (b d-a e)^3 \log (d+e x)+b^4 \left (6 d^2 e^2 x^2+9 d^3 e x-3 d^4-2 d e^3 x^3+e^4 x^4\right )\right )}{3 e^5 (a+b x) (d+e x)} \]

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]*(12*a^3*b*d*e^3 - 3*a^4*e^4 + 18*a^2*b^2*e^2*(-d^2 + d*e*x + e^2*x^2) + 6*a*b^3*e*(2*d^3 -
4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3) + b^4*(-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) - 12*b*(
b*d - a*e)^3*(d + e*x)*Log[d + e*x]))/(3*e^5*(a + b*x)*(d + e*x))

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Maple [A]  time = 0.015, size = 327, normalized size = 1.4 \begin{align*}{\frac{{x}^{4}{b}^{4}{e}^{4}+6\,{x}^{3}a{b}^{3}{e}^{4}-2\,{x}^{3}{b}^{4}d{e}^{3}+12\,\ln \left ( ex+d \right ) x{a}^{3}b{e}^{4}-36\,\ln \left ( ex+d \right ) x{a}^{2}{b}^{2}d{e}^{3}+36\,\ln \left ( ex+d \right ) xa{b}^{3}{d}^{2}{e}^{2}-12\,\ln \left ( ex+d \right ) x{b}^{4}{d}^{3}e+18\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-18\,{x}^{2}a{b}^{3}d{e}^{3}+6\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+12\,\ln \left ( ex+d \right ){a}^{3}bd{e}^{3}-36\,\ln \left ( ex+d \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}+36\,\ln \left ( ex+d \right ) a{b}^{3}{d}^{3}e-12\,\ln \left ( ex+d \right ){b}^{4}{d}^{4}+18\,x{a}^{2}{b}^{2}d{e}^{3}-24\,xa{b}^{3}{d}^{2}{e}^{2}+9\,x{b}^{4}{d}^{3}e-3\,{a}^{4}{e}^{4}+12\,d{e}^{3}{a}^{3}b-18\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}+12\,a{b}^{3}{d}^{3}e-3\,{b}^{4}{d}^{4}}{3\, \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]

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

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54662, size = 540, normalized size = 2.26 \begin{align*} \frac{b^{4} e^{4} x^{4} - 3 \, b^{4} d^{4} + 12 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} - 2 \,{\left (b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 6 \,{\left (b^{4} d^{2} e^{2} - 3 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 3 \,{\left (3 \, b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3}\right )} x - 12 \,{\left (b^{4} d^{4} - 3 \, a b^{3} d^{3} e + 3 \, a^{2} b^{2} d^{2} e^{2} - a^{3} b d e^{3} +{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (e x + d\right )}{3 \,{\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]

1/3*(b^4*e^4*x^4 - 3*b^4*d^4 + 12*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 3*a^4*e^4 - 2*(b^4*d*e^3
- 3*a*b^3*e^4)*x^3 + 6*(b^4*d^2*e^2 - 3*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 + 3*(3*b^4*d^3*e - 8*a*b^3*d^2*e^2 +
6*a^2*b^2*d*e^3)*x - 12*(b^4*d^4 - 3*a*b^3*d^3*e + 3*a^2*b^2*d^2*e^2 - a^3*b*d*e^3 + (b^4*d^3*e - 3*a*b^3*d^2
*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*log(e*x + d))/(e^6*x + d*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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15136, size = 362, normalized size = 1.51 \begin{align*} -4 \,{\left (b^{4} d^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, a b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - a^{3} 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}{3} \,{\left (b^{4} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) - 3 \, b^{4} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 9 \, b^{4} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, a b^{3} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) - 24 \, a b^{3} d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} x e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} - \frac{{\left (b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + a^{4} 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^4*d^3*sgn(b*x + a) - 3*a*b^3*d^2*e*sgn(b*x + a) + 3*a^2*b^2*d*e^2*sgn(b*x + a) - a^3*b*e^3*sgn(b*x + a))
*e^(-5)*log(abs(x*e + d)) + 1/3*(b^4*x^3*e^4*sgn(b*x + a) - 3*b^4*d*x^2*e^3*sgn(b*x + a) + 9*b^4*d^2*x*e^2*sgn
(b*x + a) + 6*a*b^3*x^2*e^4*sgn(b*x + a) - 24*a*b^3*d*x*e^3*sgn(b*x + a) + 18*a^2*b^2*x*e^4*sgn(b*x + a))*e^(-
6) - (b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b
*x + a) + a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)