3.1690 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{9/2}} \, dx\)

Optimal. Leaf size=204 \[ \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) (d+e x)^{3/2}}-\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{7 e^4 (a+b x) (d+e x)^{7/2}}-\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {646, 43} \[ -\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}+\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) (d+e x)^{3/2}}-\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{7 e^4 (a+b x) (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 119, normalized size = 0.58 \[ -\frac {2 \sqrt {(a+b x)^2} \left (5 a^3 e^3+3 a^2 b e^2 (2 d+7 e x)+a b^2 e \left (8 d^2+28 d e x+35 e^2 x^2\right )+b^3 \left (16 d^3+56 d^2 e x+70 d e^2 x^2+35 e^3 x^3\right )\right )}{35 e^4 (a+b x) (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(5*a^3*e^3 + 3*a^2*b*e^2*(2*d + 7*e*x) + a*b^2*e*(8*d^2 + 28*d*e*x + 35*e^2*x^2) + b^3*(
16*d^3 + 56*d^2*e*x + 70*d*e^2*x^2 + 35*e^3*x^3)))/(35*e^4*(a + b*x)*(d + e*x)^(7/2))

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fricas [A]  time = 0.76, size = 158, normalized size = 0.77 \[ -\frac {2 \, {\left (35 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 8 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 35 \, {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (8 \, b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{35 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]

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)^(9/2),x, algorithm="fricas")

[Out]

-2/35*(35*b^3*e^3*x^3 + 16*b^3*d^3 + 8*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 5*a^3*e^3 + 35*(2*b^3*d*e^2 + a*b^2*e^3)*
x^2 + 7*(8*b^3*d^2*e + 4*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*sqrt(e*x + d)/(e^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 +
4*d^3*e^5*x + d^4*e^4)

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giac [A]  time = 0.21, size = 194, normalized size = 0.95 \[ -\frac {2 \, {\left (35 \, {\left (x e + d\right )}^{3} b^{3} \mathrm {sgn}\left (b x + a\right ) - 35 \, {\left (x e + d\right )}^{2} b^{3} d \mathrm {sgn}\left (b x + a\right ) + 21 \, {\left (x e + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left (x e + d\right )}^{2} a b^{2} e \mathrm {sgn}\left (b x + a\right ) - 42 \, {\left (x e + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + 15 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 21 \, {\left (x e + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) - 15 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{35 \, {\left (x e + d\right )}^{\frac {7}{2}}} \]

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)^(9/2),x, algorithm="giac")

[Out]

-2/35*(35*(x*e + d)^3*b^3*sgn(b*x + a) - 35*(x*e + d)^2*b^3*d*sgn(b*x + a) + 21*(x*e + d)*b^3*d^2*sgn(b*x + a)
- 5*b^3*d^3*sgn(b*x + a) + 35*(x*e + d)^2*a*b^2*e*sgn(b*x + a) - 42*(x*e + d)*a*b^2*d*e*sgn(b*x + a) + 15*a*b
^2*d^2*e*sgn(b*x + a) + 21*(x*e + d)*a^2*b*e^2*sgn(b*x + a) - 15*a^2*b*d*e^2*sgn(b*x + a) + 5*a^3*e^3*sgn(b*x
+ a))*e^(-4)/(x*e + d)^(7/2)

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maple [A]  time = 0.05, size = 132, normalized size = 0.65 \[ -\frac {2 \left (35 b^{3} e^{3} x^{3}+35 a \,b^{2} e^{3} x^{2}+70 b^{3} d \,e^{2} x^{2}+21 a^{2} b \,e^{3} x +28 a \,b^{2} d \,e^{2} x +56 b^{3} d^{2} e x +5 a^{3} e^{3}+6 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (e x +d \right )^{\frac {7}{2}} \left (b x +a \right )^{3} e^{4}} \]

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)^(9/2),x)

[Out]

-2/35/(e*x+d)^(7/2)*(35*b^3*e^3*x^3+35*a*b^2*e^3*x^2+70*b^3*d*e^2*x^2+21*a^2*b*e^3*x+28*a*b^2*d*e^2*x+56*b^3*d
^2*e*x+5*a^3*e^3+6*a^2*b*d*e^2+8*a*b^2*d^2*e+16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3

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maxima [A]  time = 1.16, size = 147, normalized size = 0.72 \[ -\frac {2 \, {\left (35 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 8 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 35 \, {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (8 \, b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )}}{35 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )} \sqrt {e x + d}} \]

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)^(9/2),x, algorithm="maxima")

[Out]

-2/35*(35*b^3*e^3*x^3 + 16*b^3*d^3 + 8*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 5*a^3*e^3 + 35*(2*b^3*d*e^2 + a*b^2*e^3)*
x^2 + 7*(8*b^3*d^2*e + 4*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)/((e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)*sqrt(e
*x + d))

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mupad [B]  time = 1.33, size = 239, normalized size = 1.17 \[ -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {10\,a^3\,e^3+12\,a^2\,b\,d\,e^2+16\,a\,b^2\,d^2\,e+32\,b^3\,d^3}{35\,b\,e^7}+\frac {2\,x\,\left (3\,a^2\,e^2+4\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^6}+\frac {2\,b^2\,x^3}{e^4}+\frac {2\,b\,x^2\,\left (a\,e+2\,b\,d\right )}{e^5}\right )}{x^4\,\sqrt {d+e\,x}+\frac {a\,d^3\,\sqrt {d+e\,x}}{b\,e^3}+\frac {x^3\,\left (35\,a\,e^7+105\,b\,d\,e^6\right )\,\sqrt {d+e\,x}}{35\,b\,e^7}+\frac {3\,d\,x^2\,\left (a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}+\frac {d^2\,x\,\left (3\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^3}} \]

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)^(9/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((10*a^3*e^3 + 32*b^3*d^3 + 16*a*b^2*d^2*e + 12*a^2*b*d*e^2)/(35*b*e^7) + (2
*x*(3*a^2*e^2 + 8*b^2*d^2 + 4*a*b*d*e))/(5*e^6) + (2*b^2*x^3)/e^4 + (2*b*x^2*(a*e + 2*b*d))/e^5))/(x^4*(d + e*
x)^(1/2) + (a*d^3*(d + e*x)^(1/2))/(b*e^3) + (x^3*(35*a*e^7 + 105*b*d*e^6)*(d + e*x)^(1/2))/(35*b*e^7) + (3*d*
x^2*(a*e + b*d)*(d + e*x)^(1/2))/(b*e^2) + (d^2*x*(3*a*e + b*d)*(d + e*x)^(1/2))/(b*e^3))

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

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)**(9/2),x)

[Out]

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

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