3.1567 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^8} \, dx\)

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

[Out]

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

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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} \[ -\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^4 (a+b x) (d+e x)^4}+\frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{5 e^4 (a+b x) (d+e x)^5}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^4 (a+b x) (d+e x)^6}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{7 e^4 (a+b x) (d+e x)^7} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^8,x]

[Out]

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

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)^8} \, 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)^8} \, 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)^8}+\frac {3 b^4 (b d-a e)^2}{e^3 (d+e x)^7}-\frac {3 b^5 (b d-a e)}{e^3 (d+e x)^6}+\frac {b^6}{e^3 (d+e x)^5}\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}}{7 e^4 (a+b x) (d+e x)^7}-\frac {b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x) (d+e x)^6}+\frac {3 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^5}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^4 (a+b x) (d+e x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 112, normalized size = 0.56 \[ -\frac {\sqrt {(a+b x)^2} \left (20 a^3 e^3+10 a^2 b e^2 (d+7 e x)+4 a b^2 e \left (d^2+7 d e x+21 e^2 x^2\right )+b^3 \left (d^3+7 d^2 e x+21 d e^2 x^2+35 e^3 x^3\right )\right )}{140 e^4 (a+b x) (d+e x)^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^8,x]

[Out]

-1/140*(Sqrt[(a + b*x)^2]*(20*a^3*e^3 + 10*a^2*b*e^2*(d + 7*e*x) + 4*a*b^2*e*(d^2 + 7*d*e*x + 21*e^2*x^2) + b^
3*(d^3 + 7*d^2*e*x + 21*d*e^2*x^2 + 35*e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^7)

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fricas [A]  time = 0.79, size = 182, normalized size = 0.91 \[ -\frac {35 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 4 \, a b^{2} d^{2} e + 10 \, a^{2} b d e^{2} + 20 \, a^{3} e^{3} + 21 \, {\left (b^{3} d e^{2} + 4 \, a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 10 \, a^{2} b e^{3}\right )} x}{140 \, {\left (e^{11} x^{7} + 7 \, d e^{10} x^{6} + 21 \, d^{2} e^{9} x^{5} + 35 \, d^{3} e^{8} x^{4} + 35 \, d^{4} e^{7} x^{3} + 21 \, d^{5} e^{6} x^{2} + 7 \, d^{6} e^{5} x + d^{7} e^{4}\right )}} \]

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)^8,x, algorithm="fricas")

[Out]

-1/140*(35*b^3*e^3*x^3 + b^3*d^3 + 4*a*b^2*d^2*e + 10*a^2*b*d*e^2 + 20*a^3*e^3 + 21*(b^3*d*e^2 + 4*a*b^2*e^3)*
x^2 + 7*(b^3*d^2*e + 4*a*b^2*d*e^2 + 10*a^2*b*e^3)*x)/(e^11*x^7 + 7*d*e^10*x^6 + 21*d^2*e^9*x^5 + 35*d^3*e^8*x
^4 + 35*d^4*e^7*x^3 + 21*d^5*e^6*x^2 + 7*d^6*e^5*x + d^7*e^4)

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giac [A]  time = 0.19, size = 169, normalized size = 0.84 \[ -\frac {{\left (35 \, b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 21 \, b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 84 \, a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 70 \, a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{140 \, {\left (x e + d\right )}^{7}} \]

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)^8,x, algorithm="giac")

[Out]

-1/140*(35*b^3*x^3*e^3*sgn(b*x + a) + 21*b^3*d*x^2*e^2*sgn(b*x + a) + 7*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn
(b*x + a) + 84*a*b^2*x^2*e^3*sgn(b*x + a) + 28*a*b^2*d*x*e^2*sgn(b*x + a) + 4*a*b^2*d^2*e*sgn(b*x + a) + 70*a^
2*b*x*e^3*sgn(b*x + a) + 10*a^2*b*d*e^2*sgn(b*x + a) + 20*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^7

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maple [A]  time = 0.04, size = 131, normalized size = 0.66 \[ -\frac {\left (35 b^{3} e^{3} x^{3}+84 a \,b^{2} e^{3} x^{2}+21 b^{3} d \,e^{2} x^{2}+70 a^{2} b \,e^{3} x +28 a \,b^{2} d \,e^{2} x +7 b^{3} d^{2} e x +20 a^{3} e^{3}+10 a^{2} b d \,e^{2}+4 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (e x +d \right )^{7} \left (b x +a \right )^{3} e^{4}} \]

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)^8,x)

[Out]

-1/140/e^4*(35*b^3*e^3*x^3+84*a*b^2*e^3*x^2+21*b^3*d*e^2*x^2+70*a^2*b*e^3*x+28*a*b^2*d*e^2*x+7*b^3*d^2*e*x+20*
a^3*e^3+10*a^2*b*d*e^2+4*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^7/(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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^8,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 = 0.65, size = 284, normalized size = 1.42 \[ \frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{5\,e^4}+\frac {b^3\,d}{5\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{6\,e^4}+\frac {d\,\left (\frac {b^3\,d}{6\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{6\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {a^3}{7\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{7\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{7\,e}-\frac {b^3\,d}{7\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \]

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)^8,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{8}}\, dx \]

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)**8,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**8, x)

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