3.1566 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^7} \, dx\)

Optimal. Leaf size=143 \[ \frac {b^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{60 (d+e x)^4 (b d-a e)^3}+\frac {b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{15 (d+e x)^5 (b d-a e)^2}+\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{6 (d+e x)^6 (b d-a e)} \]

[Out]

1/6*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)/(e*x+d)^6+1/15*b*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+
b*d)^2/(e*x+d)^5+1/60*b^2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(-a*e+b*d)^3/(e*x+d)^4

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Rubi [A]  time = 0.09, antiderivative size = 200, normalized size of antiderivative = 1.40, 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} \[ -\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^3}+\frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{4 e^4 (a+b x) (d+e x)^4}-\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^4 (a+b x) (d+e x)^5}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{6 e^4 (a+b x) (d+e x)^6} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^7,x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 112, normalized size = 0.78 \[ -\frac {\sqrt {(a+b x)^2} \left (10 a^3 e^3+6 a^2 b e^2 (d+6 e x)+3 a b^2 e \left (d^2+6 d e x+15 e^2 x^2\right )+b^3 \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^4 (a+b x) (d+e x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^7,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(10*a^3*e^3 + 6*a^2*b*e^2*(d + 6*e*x) + 3*a*b^2*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + b^3*
(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^6)

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fricas [A]  time = 1.11, size = 171, normalized size = 1.20 \[ -\frac {20 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 10 \, a^{3} e^{3} + 15 \, {\left (b^{3} d e^{2} + 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e + 3 \, a b^{2} d e^{2} + 6 \, a^{2} b e^{3}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]

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)^7,x, algorithm="fricas")

[Out]

-1/60*(20*b^3*e^3*x^3 + b^3*d^3 + 3*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 10*a^3*e^3 + 15*(b^3*d*e^2 + 3*a*b^2*e^3)*x^
2 + 6*(b^3*d^2*e + 3*a*b^2*d*e^2 + 6*a^2*b*e^3)*x)/(e^10*x^6 + 6*d*e^9*x^5 + 15*d^2*e^8*x^4 + 20*d^3*e^7*x^3 +
15*d^4*e^6*x^2 + 6*d^5*e^5*x + d^6*e^4)

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giac [A]  time = 0.17, size = 169, normalized size = 1.18 \[ -\frac {{\left (20 \, b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 45 \, a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 36 \, a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{60 \, {\left (x e + d\right )}^{6}} \]

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)^7,x, algorithm="giac")

[Out]

-1/60*(20*b^3*x^3*e^3*sgn(b*x + a) + 15*b^3*d*x^2*e^2*sgn(b*x + a) + 6*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn(
b*x + a) + 45*a*b^2*x^2*e^3*sgn(b*x + a) + 18*a*b^2*d*x*e^2*sgn(b*x + a) + 3*a*b^2*d^2*e*sgn(b*x + a) + 36*a^2
*b*x*e^3*sgn(b*x + a) + 6*a^2*b*d*e^2*sgn(b*x + a) + 10*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^6

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maple [A]  time = 0.05, size = 131, normalized size = 0.92 \[ -\frac {\left (20 b^{3} e^{3} x^{3}+45 a \,b^{2} e^{3} x^{2}+15 b^{3} d \,e^{2} x^{2}+36 a^{2} b \,e^{3} x +18 a \,b^{2} d \,e^{2} x +6 b^{3} d^{2} e x +10 a^{3} e^{3}+6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (e x +d \right )^{6} \left (b x +a \right )^{3} e^{4}} \]

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)^7,x)

[Out]

-1/60/e^4*(20*b^3*e^3*x^3+45*a*b^2*e^3*x^2+15*b^3*d*e^2*x^2+36*a^2*b*e^3*x+18*a*b^2*d*e^2*x+6*b^3*d^2*e*x+10*a
^3*e^3+6*a^2*b*d*e^2+3*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^6/(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)^7,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 = 0.64, size = 284, normalized size = 1.99 \[ \frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{4\,e^4}+\frac {b^3\,d}{4\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{5\,e^4}+\frac {d\,\left (\frac {b^3\,d}{5\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{5\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {a^3}{6\,e}-\frac {d\,\left (\frac {a^2\,b}{2\,e}-\frac {d\,\left (\frac {a\,b^2}{2\,e}-\frac {b^3\,d}{6\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \]

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)^7,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{7}}\, dx \]

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)**7,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**7, x)

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