3.1987 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{12}} \, dx\)

Optimal. Leaf size=254 \[ -\frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}+\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^5 (a+b x) (d+e x)^8}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^5 (a+b x) (d+e x)^9}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^5 (a+b x) (d+e x)^{10}}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{11 e^5 (a+b x) (d+e x)^{11}} \]

[Out]

-((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*e^5*(a + b*x)*(d + e*x)^11) + (2*b*(b*d - a*e)^3*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^10) - (2*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e
^5*(a + b*x)*(d + e*x)^9) + (b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^5*(a + b*x)*(d + e*x)^8) - (b
^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)*(d + e*x)^7)

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Rubi [A]  time = 0.133396, antiderivative size = 254, 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} \[ -\frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}+\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^5 (a+b x) (d+e x)^8}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^5 (a+b x) (d+e x)^9}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^5 (a+b x) (d+e x)^{10}}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{11 e^5 (a+b x) (d+e x)^{11}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^12,x]

[Out]

-((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*e^5*(a + b*x)*(d + e*x)^11) + (2*b*(b*d - a*e)^3*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^10) - (2*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e
^5*(a + b*x)*(d + e*x)^9) + (b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^5*(a + b*x)*(d + e*x)^8) - (b
^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)*(d + e*x)^7)

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)^{12}} \, 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)^{12}} \, 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)^{12}} \, 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)^{12}}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^{11}}+\frac{6 b^2 (b d-a e)^2}{e^4 (d+e x)^{10}}-\frac{4 b^3 (b d-a e)}{e^4 (d+e x)^9}+\frac{b^4}{e^4 (d+e x)^8}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{11 e^5 (a+b x) (d+e x)^{11}}+\frac{2 b (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{10}}-\frac{2 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^9}+\frac{b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^8}-\frac{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}\\ \end{align*}

Mathematica [A]  time = 0.0743466, size = 162, normalized size = 0.64 \[ -\frac{\sqrt{(a+b x)^2} \left (28 a^2 b^2 e^2 \left (d^2+11 d e x+55 e^2 x^2\right )+84 a^3 b e^3 (d+11 e x)+210 a^4 e^4+7 a b^3 e \left (11 d^2 e x+d^3+55 d e^2 x^2+165 e^3 x^3\right )+b^4 \left (55 d^2 e^2 x^2+11 d^3 e x+d^4+165 d e^3 x^3+330 e^4 x^4\right )\right )}{2310 e^5 (a+b x) (d+e x)^{11}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^12,x]

[Out]

-(Sqrt[(a + b*x)^2]*(210*a^4*e^4 + 84*a^3*b*e^3*(d + 11*e*x) + 28*a^2*b^2*e^2*(d^2 + 11*d*e*x + 55*e^2*x^2) +
7*a*b^3*e*(d^3 + 11*d^2*e*x + 55*d*e^2*x^2 + 165*e^3*x^3) + b^4*(d^4 + 11*d^3*e*x + 55*d^2*e^2*x^2 + 165*d*e^3
*x^3 + 330*e^4*x^4)))/(2310*e^5*(a + b*x)*(d + e*x)^11)

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Maple [A]  time = 0.01, size = 201, normalized size = 0.8 \begin{align*} -{\frac{330\,{x}^{4}{b}^{4}{e}^{4}+1155\,{x}^{3}a{b}^{3}{e}^{4}+165\,{x}^{3}{b}^{4}d{e}^{3}+1540\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+385\,{x}^{2}a{b}^{3}d{e}^{3}+55\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+924\,x{a}^{3}b{e}^{4}+308\,x{a}^{2}{b}^{2}d{e}^{3}+77\,xa{b}^{3}{d}^{2}{e}^{2}+11\,x{b}^{4}{d}^{3}e+210\,{a}^{4}{e}^{4}+84\,d{e}^{3}{a}^{3}b+28\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}+7\,a{b}^{3}{d}^{3}e+{b}^{4}{d}^{4}}{2310\,{e}^{5} \left ( ex+d \right ) ^{11} \left ( bx+a \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)^12,x)

[Out]

-1/2310/e^5*(330*b^4*e^4*x^4+1155*a*b^3*e^4*x^3+165*b^4*d*e^3*x^3+1540*a^2*b^2*e^4*x^2+385*a*b^3*d*e^3*x^2+55*
b^4*d^2*e^2*x^2+924*a^3*b*e^4*x+308*a^2*b^2*d*e^3*x+77*a*b^3*d^2*e^2*x+11*b^4*d^3*e*x+210*a^4*e^4+84*a^3*b*d*e
^3+28*a^2*b^2*d^2*e^2+7*a*b^3*d^3*e+b^4*d^4)*((b*x+a)^2)^(3/2)/(e*x+d)^11/(b*x+a)^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)^12,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53013, size = 636, normalized size = 2.5 \begin{align*} -\frac{330 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + 7 \, a b^{3} d^{3} e + 28 \, a^{2} b^{2} d^{2} e^{2} + 84 \, a^{3} b d e^{3} + 210 \, a^{4} e^{4} + 165 \,{\left (b^{4} d e^{3} + 7 \, a b^{3} e^{4}\right )} x^{3} + 55 \,{\left (b^{4} d^{2} e^{2} + 7 \, a b^{3} d e^{3} + 28 \, a^{2} b^{2} e^{4}\right )} x^{2} + 11 \,{\left (b^{4} d^{3} e + 7 \, a b^{3} d^{2} e^{2} + 28 \, a^{2} b^{2} d e^{3} + 84 \, a^{3} b e^{4}\right )} x}{2310 \,{\left (e^{16} x^{11} + 11 \, d e^{15} x^{10} + 55 \, d^{2} e^{14} x^{9} + 165 \, d^{3} e^{13} x^{8} + 330 \, d^{4} e^{12} x^{7} + 462 \, d^{5} e^{11} x^{6} + 462 \, d^{6} e^{10} x^{5} + 330 \, d^{7} e^{9} x^{4} + 165 \, d^{8} e^{8} x^{3} + 55 \, d^{9} e^{7} x^{2} + 11 \, d^{10} e^{6} x + d^{11} 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)^12,x, algorithm="fricas")

[Out]

-1/2310*(330*b^4*e^4*x^4 + b^4*d^4 + 7*a*b^3*d^3*e + 28*a^2*b^2*d^2*e^2 + 84*a^3*b*d*e^3 + 210*a^4*e^4 + 165*(
b^4*d*e^3 + 7*a*b^3*e^4)*x^3 + 55*(b^4*d^2*e^2 + 7*a*b^3*d*e^3 + 28*a^2*b^2*e^4)*x^2 + 11*(b^4*d^3*e + 7*a*b^3
*d^2*e^2 + 28*a^2*b^2*d*e^3 + 84*a^3*b*e^4)*x)/(e^16*x^11 + 11*d*e^15*x^10 + 55*d^2*e^14*x^9 + 165*d^3*e^13*x^
8 + 330*d^4*e^12*x^7 + 462*d^5*e^11*x^6 + 462*d^6*e^10*x^5 + 330*d^7*e^9*x^4 + 165*d^8*e^8*x^3 + 55*d^9*e^7*x^
2 + 11*d^10*e^6*x + d^11*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)**12,x)

[Out]

Timed out

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Giac [A]  time = 1.12421, size = 356, normalized size = 1.4 \begin{align*} -\frac{{\left (330 \, b^{4} x^{4} e^{4} \mathrm{sgn}\left (b x + a\right ) + 165 \, b^{4} d x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 55 \, b^{4} d^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 11 \, b^{4} d^{3} x e \mathrm{sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) + 1155 \, a b^{3} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 385 \, a b^{3} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 77 \, a b^{3} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 7 \, a b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 1540 \, a^{2} b^{2} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 308 \, a^{2} b^{2} d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 28 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 924 \, a^{3} b x e^{4} \mathrm{sgn}\left (b x + a\right ) + 84 \, a^{3} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + 210 \, a^{4} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{2310 \,{\left (x e + d\right )}^{11}} \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)^12,x, algorithm="giac")

[Out]

-1/2310*(330*b^4*x^4*e^4*sgn(b*x + a) + 165*b^4*d*x^3*e^3*sgn(b*x + a) + 55*b^4*d^2*x^2*e^2*sgn(b*x + a) + 11*
b^4*d^3*x*e*sgn(b*x + a) + b^4*d^4*sgn(b*x + a) + 1155*a*b^3*x^3*e^4*sgn(b*x + a) + 385*a*b^3*d*x^2*e^3*sgn(b*
x + a) + 77*a*b^3*d^2*x*e^2*sgn(b*x + a) + 7*a*b^3*d^3*e*sgn(b*x + a) + 1540*a^2*b^2*x^2*e^4*sgn(b*x + a) + 30
8*a^2*b^2*d*x*e^3*sgn(b*x + a) + 28*a^2*b^2*d^2*e^2*sgn(b*x + a) + 924*a^3*b*x*e^4*sgn(b*x + a) + 84*a^3*b*d*e
^3*sgn(b*x + a) + 210*a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^11