3.13.70 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^9} \, dx\)

Optimal. Leaf size=200 \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^4 (a+b x) (d+e x)^6}-\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{7 e^4 (a+b x) (d+e x)^7}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{8 e^4 (a+b x) (d+e x)^8}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^5} \]

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Rubi [A]  time = 0.09, antiderivative size = 200, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^5}+\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^4 (a+b x) (d+e x)^6}-\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{7 e^4 (a+b x) (d+e x)^7}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{8 e^4 (a+b x) (d+e x)^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^9,x]

[Out]

((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*e^4*(a + b*x)*(d + e*x)^8) - (3*b*(b*d - a*e)^2*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(7*e^4*(a + b*x)*(d + e*x)^7) + (b^2*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^4*(a +
b*x)*(d + e*x)^6) - (b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^4*(a + b*x)*(d + e*x)^5)

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 112, normalized size = 0.56 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (35 a^3 e^3+15 a^2 b e^2 (d+8 e x)+5 a b^2 e \left (d^2+8 d e x+28 e^2 x^2\right )+b^3 \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )\right )}{280 e^4 (a+b x) (d+e x)^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^9,x]

[Out]

-1/280*(Sqrt[(a + b*x)^2]*(35*a^3*e^3 + 15*a^2*b*e^2*(d + 8*e*x) + 5*a*b^2*e*(d^2 + 8*d*e*x + 28*e^2*x^2) + b^
3*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^8)

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IntegrateAlgebraic [F]  time = 180.34, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^9,x]

[Out]

$Aborted

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fricas [A]  time = 0.41, size = 193, normalized size = 0.96 \begin {gather*} -\frac {56 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 5 \, a b^{2} d^{2} e + 15 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} + 28 \, {\left (b^{3} d e^{2} + 5 \, a b^{2} e^{3}\right )} x^{2} + 8 \, {\left (b^{3} d^{2} e + 5 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x}{280 \, {\left (e^{12} x^{8} + 8 \, d e^{11} x^{7} + 28 \, d^{2} e^{10} x^{6} + 56 \, d^{3} e^{9} x^{5} + 70 \, d^{4} e^{8} x^{4} + 56 \, d^{5} e^{7} x^{3} + 28 \, d^{6} e^{6} x^{2} + 8 \, d^{7} e^{5} x + d^{8} e^{4}\right )}} \end {gather*}

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)^9,x, algorithm="fricas")

[Out]

-1/280*(56*b^3*e^3*x^3 + b^3*d^3 + 5*a*b^2*d^2*e + 15*a^2*b*d*e^2 + 35*a^3*e^3 + 28*(b^3*d*e^2 + 5*a*b^2*e^3)*
x^2 + 8*(b^3*d^2*e + 5*a*b^2*d*e^2 + 15*a^2*b*e^3)*x)/(e^12*x^8 + 8*d*e^11*x^7 + 28*d^2*e^10*x^6 + 56*d^3*e^9*
x^5 + 70*d^4*e^8*x^4 + 56*d^5*e^7*x^3 + 28*d^6*e^6*x^2 + 8*d^7*e^5*x + d^8*e^4)

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giac [A]  time = 0.19, size = 169, normalized size = 0.84 \begin {gather*} -\frac {{\left (56 \, b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 140 \, a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 40 \, a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{280 \, {\left (x e + d\right )}^{8}} \end {gather*}

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)^9,x, algorithm="giac")

[Out]

-1/280*(56*b^3*x^3*e^3*sgn(b*x + a) + 28*b^3*d*x^2*e^2*sgn(b*x + a) + 8*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn
(b*x + a) + 140*a*b^2*x^2*e^3*sgn(b*x + a) + 40*a*b^2*d*x*e^2*sgn(b*x + a) + 5*a*b^2*d^2*e*sgn(b*x + a) + 120*
a^2*b*x*e^3*sgn(b*x + a) + 15*a^2*b*d*e^2*sgn(b*x + a) + 35*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^8

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maple [A]  time = 0.05, size = 131, normalized size = 0.66 \begin {gather*} -\frac {\left (56 b^{3} e^{3} x^{3}+140 a \,b^{2} e^{3} x^{2}+28 b^{3} d \,e^{2} x^{2}+120 a^{2} b \,e^{3} x +40 a \,b^{2} d \,e^{2} x +8 b^{3} d^{2} e x +35 a^{3} e^{3}+15 a^{2} b d \,e^{2}+5 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (e x +d \right )^{8} \left (b x +a \right )^{3} e^{4}} \end {gather*}

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)^9,x)

[Out]

-1/280/e^4*(56*b^3*e^3*x^3+140*a*b^2*e^3*x^2+28*b^3*d*e^2*x^2+120*a^2*b*e^3*x+40*a*b^2*d*e^2*x+8*b^3*d^2*e*x+3
5*a^3*e^3+15*a^2*b*d*e^2+5*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^8/(b*x+a)^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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)^9,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 = 0.66, size = 284, normalized size = 1.42 \begin {gather*} \frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{6\,e^4}+\frac {b^3\,d}{6\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{7\,e^4}+\frac {d\,\left (\frac {b^3\,d}{7\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{7\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {a^3}{8\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{8\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{8\,e}-\frac {b^3\,d}{8\,e^2}\right )}{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 )}^8}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^9,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{9}}\, dx \end {gather*}

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)**9,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**9, x)

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