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

Optimal. Leaf size=202 \[ \frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) \sqrt {d+e x}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)^{3/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^4 (a+b x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 202, 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} \begin {gather*} \frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^4 (a+b x)}+\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) \sqrt {d+e x}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)^{3/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^4 (a+b x) (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 118, normalized size = 0.58 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (a^3 e^3+a^2 b e^2 (2 d+5 e x)+a b^2 e \left (8 d^2+20 d e x+15 e^2 x^2\right )-\left (b^3 \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )\right )}{5 e^4 (a+b x) (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(a^3*e^3 + a^2*b*e^2*(2*d + 5*e*x) + a*b^2*e*(8*d^2 + 20*d*e*x + 15*e^2*x^2) - b^3*(16*d
^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3)))/(5*e^4*(a + b*x)*(d + e*x)^(5/2))

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IntegrateAlgebraic [A]  time = 24.83, size = 159, normalized size = 0.79 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-a^3 e^3-5 a^2 b e^2 (d+e x)+3 a^2 b d e^2-3 a b^2 d^2 e-15 a b^2 e (d+e x)^2+10 a b^2 d e (d+e x)+b^3 d^3-5 b^3 d^2 (d+e x)+5 b^3 (d+e x)^3+15 b^3 d (d+e x)^2\right )}{5 e^3 (d+e x)^{5/2} (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a*e + b*e*x)^2/e^2]*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3 - 5*b^3*d^2*(d + e*x) + 10*a*b
^2*d*e*(d + e*x) - 5*a^2*b*e^2*(d + e*x) + 15*b^3*d*(d + e*x)^2 - 15*a*b^2*e*(d + e*x)^2 + 5*b^3*(d + e*x)^3))
/(5*e^3*(d + e*x)^(5/2)*(a*e + b*e*x))

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fricas [A]  time = 0.40, size = 148, normalized size = 0.73 \begin {gather*} \frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} 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)^(7/2),x, algorithm="fricas")

[Out]

2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 +
5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2*b*e^3)*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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giac [A]  time = 0.30, size = 196, normalized size = 0.97 \begin {gather*} 2 \, \sqrt {x e + d} b^{3} e^{\left (-4\right )} \mathrm {sgn}\left (b x + a\right ) + \frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} b^{3} d \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 15 \, {\left (x e + d\right )}^{2} a b^{2} e \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (x e + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{5 \, {\left (x e + d\right )}^{\frac {5}{2}}} \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)^(7/2),x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*b^3*e^(-4)*sgn(b*x + a) + 2/5*(15*(x*e + d)^2*b^3*d*sgn(b*x + a) - 5*(x*e + d)*b^3*d^2*sgn(b*x
+ a) + b^3*d^3*sgn(b*x + a) - 15*(x*e + d)^2*a*b^2*e*sgn(b*x + a) + 10*(x*e + d)*a*b^2*d*e*sgn(b*x + a) - 3*a
*b^2*d^2*e*sgn(b*x + a) - 5*(x*e + d)*a^2*b*e^2*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x +
a))*e^(-4)/(x*e + d)^(5/2)

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maple [A]  time = 0.05, size = 131, normalized size = 0.65 \begin {gather*} -\frac {2 \left (-5 b^{3} e^{3} x^{3}+15 a \,b^{2} e^{3} x^{2}-30 b^{3} d \,e^{2} x^{2}+5 a^{2} b \,e^{3} x +20 a \,b^{2} d \,e^{2} x -40 b^{3} d^{2} e x +a^{3} e^{3}+2 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}}}{5 \left (e x +d \right )^{\frac {5}{2}} \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)^(7/2),x)

[Out]

-2/5/(e*x+d)^(5/2)*(-5*b^3*e^3*x^3+15*a*b^2*e^3*x^2-30*b^3*d*e^2*x^2+5*a^2*b*e^3*x+20*a*b^2*d*e^2*x-40*b^3*d^2
*e*x+a^3*e^3+2*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 = 1.23, size = 137, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )}}{5 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt {e x + d}} \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)^(7/2),x, algorithm="maxima")

[Out]

2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 +
5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2*b*e^3)*x)/((e^6*x^2 + 2*d*e^5*x + d^2*e^4)*sqrt(e*x + d))

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mupad [B]  time = 1.29, size = 210, normalized size = 1.04 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,a^3\,e^3+4\,a^2\,b\,d\,e^2+16\,a\,b^2\,d^2\,e-32\,b^3\,d^3}{5\,b\,e^6}+\frac {2\,x\,\left (a^2\,e^2+4\,a\,b\,d\,e-8\,b^2\,d^2\right )}{e^5}-\frac {2\,b^2\,x^3}{e^3}+\frac {6\,b\,x^2\,\left (a\,e-2\,b\,d\right )}{e^4}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (5\,a\,e^6+10\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{5\,b\,e^6}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \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)^(7/2),x)

[Out]

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

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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 )^{\frac {7}{2}}}\, 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)**(7/2),x)

[Out]

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

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