3.14.90 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

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

________________________________________________________________________________________

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

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 119, normalized size = 0.58 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (a^3 e^3+3 a^2 b e^2 (2 d+3 e x)-3 a b^2 e \left (8 d^2+12 d e x+3 e^2 x^2\right )+b^3 \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )\right )}{3 e^4 (a+b x) (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(a^3*e^3 + 3*a^2*b*e^2*(2*d + 3*e*x) - 3*a*b^2*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + b^3*(1
6*d^3 + 24*d^2*e*x + 6*d*e^2*x^2 - e^3*x^3)))/(3*e^4*(a + b*x)*(d + e*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 21.98, size = 158, normalized size = 0.77 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-a^3 e^3-9 a^2 b e^2 (d+e x)+3 a^2 b d e^2-3 a b^2 d^2 e+9 a b^2 e (d+e x)^2+18 a b^2 d e (d+e x)+b^3 d^3-9 b^3 d^2 (d+e x)+b^3 (d+e x)^3-9 b^3 d (d+e x)^2\right )}{3 e^3 (d+e x)^{3/2} (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)^(5/2),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 - 9*b^3*d^2*(d + e*x) + 18*a*b
^2*d*e*(d + e*x) - 9*a^2*b*e^2*(d + e*x) - 9*b^3*d*(d + e*x)^2 + 9*a*b^2*e*(d + e*x)^2 + b^3*(d + e*x)^3))/(3*
e^3*(d + e*x)^(3/2)*(a*e + b*e*x))

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 136, normalized size = 0.67 \begin {gather*} \frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*sqrt(e*x + d)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)

________________________________________________________________________________________

giac [A]  time = 0.22, size = 202, normalized size = 0.99 \begin {gather*} \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{3} e^{8} \mathrm {sgn}\left (b x + a\right ) - 9 \, \sqrt {x e + d} b^{3} d e^{8} \mathrm {sgn}\left (b x + a\right ) + 9 \, \sqrt {x e + d} a b^{2} e^{9} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-12\right )} - \frac {2 \, {\left (9 \, {\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 ) - 18 \, {\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 ) + 9 \, {\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 )}}{3 \, {\left (x e + d\right )}^{\frac {3}{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)^(5/2),x, algorithm="giac")

[Out]

2/3*((x*e + d)^(3/2)*b^3*e^8*sgn(b*x + a) - 9*sqrt(x*e + d)*b^3*d*e^8*sgn(b*x + a) + 9*sqrt(x*e + d)*a*b^2*e^9
*sgn(b*x + a))*e^(-12) - 2/3*(9*(x*e + d)*b^3*d^2*sgn(b*x + a) - b^3*d^3*sgn(b*x + a) - 18*(x*e + d)*a*b^2*d*e
*sgn(b*x + a) + 3*a*b^2*d^2*e*sgn(b*x + a) + 9*(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)^(3/2)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 131, normalized size = 0.64 \begin {gather*} -\frac {2 \left (-b^{3} e^{3} x^{3}-9 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+9 a^{2} b \,e^{3} x -36 a \,b^{2} d \,e^{2} x +24 b^{3} d^{2} e x +a^{3} e^{3}+6 a^{2} b d \,e^{2}-24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3 \left (e x +d \right )^{\frac {3}{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)^(5/2),x)

[Out]

-2/3/(e*x+d)^(3/2)*(-b^3*e^3*x^3-9*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+9*a^2*b*e^3*x-36*a*b^2*d*e^2*x+24*b^3*d^2*e*x
+a^3*e^3+6*a^2*b*d*e^2-24*a*b^2*d^2*e+16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3

________________________________________________________________________________________

maxima [A]  time = 1.16, size = 125, normalized size = 0.61 \begin {gather*} \frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )}}{3 \, {\left (e^{5} x + d 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)^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)/((e^5*x + d*e^4)*sqrt(e*x + d))

________________________________________________________________________________________

mupad [B]  time = 1.23, size = 184, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,a^3\,e^3+12\,a^2\,b\,d\,e^2-48\,a\,b^2\,d^2\,e+32\,b^3\,d^3}{3\,b\,e^5}+\frac {2\,x\,\left (3\,a^2\,e^2-12\,a\,b\,d\,e+8\,b^2\,d^2\right )}{e^4}-\frac {2\,b^2\,x^3}{3\,e^2}-\frac {2\,b\,x^2\,\left (3\,a\,e-2\,b\,d\right )}{e^3}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^5+3\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{3\,b\,e^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)^(5/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*a^3*e^3 + 32*b^3*d^3 - 48*a*b^2*d^2*e + 12*a^2*b*d*e^2)/(3*b*e^5) + (2*x
*(3*a^2*e^2 + 8*b^2*d^2 - 12*a*b*d*e))/e^4 - (2*b^2*x^3)/(3*e^2) - (2*b*x^2*(3*a*e - 2*b*d))/e^3))/(x^2*(d + e
*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(3*a*e^5 + 3*b*d*e^4)*(d + e*x)^(1/2))/(3*b*e^5))

________________________________________________________________________________________

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 {5}{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)**(5/2),x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**(5/2), x)

________________________________________________________________________________________