3.1980 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=246 \[ \frac{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) (d+e x)}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)^2}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^3}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{4 e^5 (a+b x) (d+e x)^4}+\frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]

[Out]

-((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*e^5*(a + b*x)*(d + e*x)^4) + (4*b*(b*d - a*e)^3*Sqrt[a^2 + 2
*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^3) - (3*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a
+ b*x)*(d + e*x)^2) + (4*b^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^4*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 = 0.135284, antiderivative size = 246, 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} \[ \frac{4 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) (d+e x)}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)^2}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^3}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{4 e^5 (a+b x) (d+e x)^4}+\frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^5,x]

[Out]

-((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*e^5*(a + b*x)*(d + e*x)^4) + (4*b*(b*d - a*e)^3*Sqrt[a^2 + 2
*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^3) - (3*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a
+ b*x)*(d + e*x)^2) + (4*b^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^4*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)^5} \, 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)^5} \, 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)^5} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^4}{e^4 (d+e x)^5}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^4}+\frac{6 b^2 (b d-a e)^2}{e^4 (d+e x)^3}-\frac{4 b^3 (b d-a e)}{e^4 (d+e x)^2}+\frac{b^4}{e^4 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4}+\frac{4 b (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^3}-\frac{3 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)^2}+\frac{4 b^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^4 \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 = 0.104711, size = 144, normalized size = 0.59 \[ \frac{\sqrt{(a+b x)^2} \left ((b d-a e) \left (a^2 b e^2 (7 d+16 e x)+3 a^3 e^3+a b^2 e \left (13 d^2+40 d e x+36 e^2 x^2\right )+b^3 \left (88 d^2 e x+25 d^3+108 d e^2 x^2+48 e^3 x^3\right )\right )+12 b^4 (d+e x)^4 \log (d+e x)\right )}{12 e^5 (a+b x) (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^5,x]

[Out]

(Sqrt[(a + b*x)^2]*((b*d - a*e)*(3*a^3*e^3 + a^2*b*e^2*(7*d + 16*e*x) + a*b^2*e*(13*d^2 + 40*d*e*x + 36*e^2*x^
2) + b^3*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3)) + 12*b^4*(d + e*x)^4*Log[d + e*x]))/(12*e^5*(a +
b*x)*(d + e*x)^4)

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Maple [A]  time = 0.013, size = 276, normalized size = 1.1 \begin{align*}{\frac{12\,\ln \left ( ex+d \right ){x}^{4}{b}^{4}{e}^{4}+48\,\ln \left ( ex+d \right ){x}^{3}{b}^{4}d{e}^{3}+72\,\ln \left ( ex+d \right ){x}^{2}{b}^{4}{d}^{2}{e}^{2}-48\,{x}^{3}a{b}^{3}{e}^{4}+48\,{x}^{3}{b}^{4}d{e}^{3}+48\,\ln \left ( ex+d \right ) x{b}^{4}{d}^{3}e-36\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-72\,{x}^{2}a{b}^{3}d{e}^{3}+108\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+12\,\ln \left ( ex+d \right ){b}^{4}{d}^{4}-16\,x{a}^{3}b{e}^{4}-24\,x{a}^{2}{b}^{2}d{e}^{3}-48\,xa{b}^{3}{d}^{2}{e}^{2}+88\,x{b}^{4}{d}^{3}e-3\,{a}^{4}{e}^{4}-4\,d{e}^{3}{a}^{3}b-6\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}-12\,a{b}^{3}{d}^{3}e+25\,{b}^{4}{d}^{4}}{12\, \left ( bx+a \right ) ^{3}{e}^{5} \left ( ex+d \right ) ^{4}} \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)^5,x)

[Out]

1/12*((b*x+a)^2)^(3/2)*(12*ln(e*x+d)*x^4*b^4*e^4+48*ln(e*x+d)*x^3*b^4*d*e^3+72*ln(e*x+d)*x^2*b^4*d^2*e^2-48*x^
3*a*b^3*e^4+48*x^3*b^4*d*e^3+48*ln(e*x+d)*x*b^4*d^3*e-36*x^2*a^2*b^2*e^4-72*x^2*a*b^3*d*e^3+108*x^2*b^4*d^2*e^
2+12*ln(e*x+d)*b^4*d^4-16*x*a^3*b*e^4-24*x*a^2*b^2*d*e^3-48*x*a*b^3*d^2*e^2+88*x*b^4*d^3*e-3*a^4*e^4-4*d*e^3*a
^3*b-6*a^2*b^2*d^2*e^2-12*a*b^3*d^3*e+25*b^4*d^4)/(b*x+a)^3/e^5/(e*x+d)^4

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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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63555, size = 544, normalized size = 2.21 \begin{align*} \frac{25 \, b^{4} d^{4} - 12 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 48 \,{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 36 \,{\left (3 \, b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 8 \,{\left (11 \, b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} - 2 \, a^{3} b e^{4}\right )} x + 12 \,{\left (b^{4} e^{4} x^{4} + 4 \, b^{4} d e^{3} x^{3} + 6 \, b^{4} d^{2} e^{2} x^{2} + 4 \, b^{4} d^{3} e x + b^{4} d^{4}\right )} \log \left (e x + d\right )}{12 \,{\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} 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)^5,x, algorithm="fricas")

[Out]

1/12*(25*b^4*d^4 - 12*a*b^3*d^3*e - 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 - 3*a^4*e^4 + 48*(b^4*d*e^3 - a*b^3*e^4)
*x^3 + 36*(3*b^4*d^2*e^2 - 2*a*b^3*d*e^3 - a^2*b^2*e^4)*x^2 + 8*(11*b^4*d^3*e - 6*a*b^3*d^2*e^2 - 3*a^2*b^2*d*
e^3 - 2*a^3*b*e^4)*x + 12*(b^4*e^4*x^4 + 4*b^4*d*e^3*x^3 + 6*b^4*d^2*e^2*x^2 + 4*b^4*d^3*e*x + b^4*d^4)*log(e*
x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^4*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)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.1319, size = 362, normalized size = 1.47 \begin{align*} b^{4} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (48 \,{\left (b^{4} d e^{2} \mathrm{sgn}\left (b x + a\right ) - a b^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} x^{3} + 36 \,{\left (3 \, b^{4} d^{2} e \mathrm{sgn}\left (b x + a\right ) - 2 \, a b^{3} d e^{2} \mathrm{sgn}\left (b x + a\right ) - a^{2} b^{2} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} x^{2} + 8 \,{\left (11 \, b^{4} d^{3} \mathrm{sgn}\left (b x + a\right ) - 6 \, 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 ) - 2 \, a^{3} b e^{3} \mathrm{sgn}\left (b x + a\right )\right )} x +{\left (25 \, b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) - 12 \, 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 ) - 3 \, a^{4} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-4\right )}}{12 \,{\left (x e + d\right )}^{4}} \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)^5,x, algorithm="giac")

[Out]

b^4*e^(-5)*log(abs(x*e + d))*sgn(b*x + a) + 1/12*(48*(b^4*d*e^2*sgn(b*x + a) - a*b^3*e^3*sgn(b*x + a))*x^3 + 3
6*(3*b^4*d^2*e*sgn(b*x + a) - 2*a*b^3*d*e^2*sgn(b*x + a) - a^2*b^2*e^3*sgn(b*x + a))*x^2 + 8*(11*b^4*d^3*sgn(b
*x + a) - 6*a*b^3*d^2*e*sgn(b*x + a) - 3*a^2*b^2*d*e^2*sgn(b*x + a) - 2*a^3*b*e^3*sgn(b*x + a))*x + (25*b^4*d^
4*sgn(b*x + a) - 12*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) - 3
*a^4*e^4*sgn(b*x + a))*e^(-1))*e^(-4)/(x*e + d)^4