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

[Out]

-1/11*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^11+2/5*b*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(
e*x+d)^10-2/3*b^2*(-a*e+b*d)^2*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^9+1/2*b^3*(-a*e+b*d)*((b*x+a)^2)^(1/2)/e^
5/(b*x+a)/(e*x+d)^8-1/7*b^4*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^7

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Rubi [A]  time = 0.13, antiderivative size = 254, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.07, size = 162, normalized size = 0.64 \[ -\frac {\sqrt {(a+b x)^2} \left (210 a^4 e^4+84 a^3 b e^3 (d+11 e x)+28 a^2 b^2 e^2 \left (d^2+11 d e x+55 e^2 x^2\right )+7 a b^3 e \left (d^3+11 d^2 e x+55 d e^2 x^2+165 e^3 x^3\right )+b^4 \left (d^4+11 d^3 e x+55 d^2 e^2 x^2+165 d e^3 x^3+330 e^4 x^4\right )\right )}{2310 e^5 (a+b x) (d+e x)^{11}} \]

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]

-1/2310*(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 + 16
5*d*e^3*x^3 + 330*e^4*x^4)))/(e^5*(a + b*x)*(d + e*x)^11)

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fricas [A]  time = 1.23, size = 291, normalized size = 1.15 \[ -\frac {330 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + 7 \, a b^{3} d^{3} e + 28 \, a^{2} b^{2} d^{2} e^{2} + 84 \, a^{3} b d e^{3} + 210 \, a^{4} e^{4} + 165 \, {\left (b^{4} d e^{3} + 7 \, a b^{3} e^{4}\right )} x^{3} + 55 \, {\left (b^{4} d^{2} e^{2} + 7 \, a b^{3} d e^{3} + 28 \, a^{2} b^{2} e^{4}\right )} x^{2} + 11 \, {\left (b^{4} d^{3} e + 7 \, a b^{3} d^{2} e^{2} + 28 \, a^{2} b^{2} d e^{3} + 84 \, a^{3} b e^{4}\right )} x}{2310 \, {\left (e^{16} x^{11} + 11 \, d e^{15} x^{10} + 55 \, d^{2} e^{14} x^{9} + 165 \, d^{3} e^{13} x^{8} + 330 \, d^{4} e^{12} x^{7} + 462 \, d^{5} e^{11} x^{6} + 462 \, d^{6} e^{10} x^{5} + 330 \, d^{7} e^{9} x^{4} + 165 \, d^{8} e^{8} x^{3} + 55 \, d^{9} e^{7} x^{2} + 11 \, d^{10} e^{6} x + d^{11} e^{5}\right )}} \]

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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giac [A]  time = 0.17, size = 264, normalized size = 1.04 \[ -\frac {{\left (330 \, b^{4} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 165 \, b^{4} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 55 \, b^{4} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 11 \, b^{4} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 1155 \, a b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 385 \, a b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 77 \, a b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 1540 \, a^{2} b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 308 \, a^{2} b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 924 \, a^{3} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 84 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 210 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{2310 \, {\left (x e + d\right )}^{11}} \]

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

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maple [A]  time = 0.06, size = 201, normalized size = 0.79 \[ -\frac {\left (330 b^{4} e^{4} x^{4}+1155 a \,b^{3} e^{4} x^{3}+165 b^{4} d \,e^{3} x^{3}+1540 a^{2} b^{2} e^{4} x^{2}+385 a \,b^{3} d \,e^{3} x^{2}+55 b^{4} d^{2} e^{2} x^{2}+924 a^{3} b \,e^{4} x +308 a^{2} b^{2} d \,e^{3} x +77 a \,b^{3} d^{2} e^{2} x +11 b^{4} d^{3} e x +210 a^{4} e^{4}+84 a^{3} b d \,e^{3}+28 a^{2} b^{2} d^{2} e^{2}+7 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2310 \left (e x +d \right )^{11} \left (b x +a \right )^{3} e^{5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.20, size = 449, normalized size = 1.77 \[ \frac {\left (\frac {-4\,a^3\,b\,e^3+6\,a^2\,b^2\,d\,e^2-4\,a\,b^3\,d^2\,e+b^4\,d^3}{10\,e^5}+\frac {d\,\left (\frac {d\,\left (\frac {b^4\,d}{10\,e^3}-\frac {b^3\,\left (4\,a\,e-b\,d\right )}{10\,e^3}\right )}{e}+\frac {b^2\,\left (6\,a^2\,e^2-4\,a\,b\,d\,e+b^2\,d^2\right )}{10\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{10}}-\frac {\left (\frac {a^4}{11\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {4\,a\,b^3}{11\,e}-\frac {b^4\,d}{11\,e^2}\right )}{e}-\frac {6\,a^2\,b^2}{11\,e}\right )}{e}+\frac {4\,a^3\,b}{11\,e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{11}}-\frac {\left (\frac {6\,a^2\,b^2\,e^2-8\,a\,b^3\,d\,e+3\,b^4\,d^2}{9\,e^5}+\frac {d\,\left (\frac {b^4\,d}{9\,e^4}-\frac {2\,b^3\,\left (2\,a\,e-b\,d\right )}{9\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^9}+\frac {\left (\frac {3\,b^4\,d-4\,a\,b^3\,e}{8\,e^5}+\frac {b^4\,d}{8\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}-\frac {b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^12,x)

[Out]

(((b^4*d^3 - 4*a^3*b*e^3 + 6*a^2*b^2*d*e^2 - 4*a*b^3*d^2*e)/(10*e^5) + (d*((d*((b^4*d)/(10*e^3) - (b^3*(4*a*e
- b*d))/(10*e^3)))/e + (b^2*(6*a^2*e^2 + b^2*d^2 - 4*a*b*d*e))/(10*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/
((a + b*x)*(d + e*x)^10) - ((a^4/(11*e) - (d*((d*((d*((4*a*b^3)/(11*e) - (b^4*d)/(11*e^2)))/e - (6*a^2*b^2)/(1
1*e)))/e + (4*a^3*b)/(11*e)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^11) - (((3*b^4*d^2 + 6*
a^2*b^2*e^2 - 8*a*b^3*d*e)/(9*e^5) + (d*((b^4*d)/(9*e^4) - (2*b^3*(2*a*e - b*d))/(9*e^4)))/e)*(a^2 + b^2*x^2 +
2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^9) + (((3*b^4*d - 4*a*b^3*e)/(8*e^5) + (b^4*d)/(8*e^5))*(a^2 + b^2*x^2 +
2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^8) - (b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*e^5*(a + b*x)*(d + e*x)^7)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

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]

Exception raised: HeuristicGCDFailed

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