3.1684 \(\int (d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=208 \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2}}{11 e^4 (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (b d-a e)}{3 e^4 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)^2}{7 e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}{5 e^4 (a+b x)} \]

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^4*(a + b*x)) + (6*b*(b*d - a*e)^2*(d + e
*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^4*(a + b*x)) - (2*b^2*(b*d - a*e)*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/(3*e^4*(a + b*x)) + (2*b^3*(d + e*x)^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*e^4*(a + b*x))

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Rubi [A]  time = 0.0651006, antiderivative size = 208, normalized size of antiderivative = 1., 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} \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2}}{11 e^4 (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (b d-a e)}{3 e^4 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)^2}{7 e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}{5 e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^4*(a + b*x)) + (6*b*(b*d - a*e)^2*(d + e
*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^4*(a + b*x)) - (2*b^2*(b*d - a*e)*(d + e*x)^(9/2)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/(3*e^4*(a + b*x)) + (2*b^3*(d + e*x)^(11/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*e^4*(a + b*x))

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]

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])

Rubi steps

\begin{align*} \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^3 (d+e x)^{3/2} \, 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 (d+e x)^{3/2}}{e^3}+\frac{3 b^4 (b d-a e)^2 (d+e x)^{5/2}}{e^3}-\frac{3 b^5 (b d-a e) (d+e x)^{7/2}}{e^3}+\frac{b^6 (d+e x)^{9/2}}{e^3}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{2 (b d-a e)^3 (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x)}+\frac{6 b (b d-a e)^2 (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x)}-\frac{2 b^2 (b d-a e) (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)}+\frac{2 b^3 (d+e x)^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}{11 e^4 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0729421, size = 120, normalized size = 0.58 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{5/2} \left (99 a^2 b e^2 (5 e x-2 d)+231 a^3 e^3+11 a b^2 e \left (8 d^2-20 d e x+35 e^2 x^2\right )+b^3 \left (40 d^2 e x-16 d^3-70 d e^2 x^2+105 e^3 x^3\right )\right )}{1155 e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(231*a^3*e^3 + 99*a^2*b*e^2*(-2*d + 5*e*x) + 11*a*b^2*e*(8*d^2 - 20*d*e*x
+ 35*e^2*x^2) + b^3*(-16*d^3 + 40*d^2*e*x - 70*d*e^2*x^2 + 105*e^3*x^3)))/(1155*e^4*(a + b*x))

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Maple [A]  time = 0.153, size = 132, normalized size = 0.6 \begin{align*}{\frac{210\,{x}^{3}{b}^{3}{e}^{3}+770\,{x}^{2}a{b}^{2}{e}^{3}-140\,{x}^{2}{b}^{3}d{e}^{2}+990\,x{a}^{2}b{e}^{3}-440\,xa{b}^{2}d{e}^{2}+80\,x{b}^{3}{d}^{2}e+462\,{a}^{3}{e}^{3}-396\,d{e}^{2}{a}^{2}b+176\,a{b}^{2}{d}^{2}e-32\,{b}^{3}{d}^{3}}{1155\,{e}^{4} \left ( bx+a \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

2/1155*(e*x+d)^(5/2)*(105*b^3*e^3*x^3+385*a*b^2*e^3*x^2-70*b^3*d*e^2*x^2+495*a^2*b*e^3*x-220*a*b^2*d*e^2*x+40*
b^3*d^2*e*x+231*a^3*e^3-198*a^2*b*d*e^2+88*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 = 1.08715, size = 292, normalized size = 1.4 \begin{align*} \frac{2 \,{\left (105 \, b^{3} e^{5} x^{5} - 16 \, b^{3} d^{5} + 88 \, a b^{2} d^{4} e - 198 \, a^{2} b d^{3} e^{2} + 231 \, a^{3} d^{2} e^{3} + 35 \,{\left (4 \, b^{3} d e^{4} + 11 \, a b^{2} e^{5}\right )} x^{4} + 5 \,{\left (b^{3} d^{2} e^{3} + 110 \, a b^{2} d e^{4} + 99 \, a^{2} b e^{5}\right )} x^{3} - 3 \,{\left (2 \, b^{3} d^{3} e^{2} - 11 \, a b^{2} d^{2} e^{3} - 264 \, a^{2} b d e^{4} - 77 \, a^{3} e^{5}\right )} x^{2} +{\left (8 \, b^{3} d^{4} e - 44 \, a b^{2} d^{3} e^{2} + 99 \, a^{2} b d^{2} e^{3} + 462 \, a^{3} d e^{4}\right )} x\right )} \sqrt{e x + d}}{1155 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

2/1155*(105*b^3*e^5*x^5 - 16*b^3*d^5 + 88*a*b^2*d^4*e - 198*a^2*b*d^3*e^2 + 231*a^3*d^2*e^3 + 35*(4*b^3*d*e^4
+ 11*a*b^2*e^5)*x^4 + 5*(b^3*d^2*e^3 + 110*a*b^2*d*e^4 + 99*a^2*b*e^5)*x^3 - 3*(2*b^3*d^3*e^2 - 11*a*b^2*d^2*e
^3 - 264*a^2*b*d*e^4 - 77*a^3*e^5)*x^2 + (8*b^3*d^4*e - 44*a*b^2*d^3*e^2 + 99*a^2*b*d^2*e^3 + 462*a^3*d*e^4)*x
)*sqrt(e*x + d)/e^4

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Fricas [A]  time = 1.54609, size = 474, normalized size = 2.28 \begin{align*} \frac{2 \,{\left (105 \, b^{3} e^{5} x^{5} - 16 \, b^{3} d^{5} + 88 \, a b^{2} d^{4} e - 198 \, a^{2} b d^{3} e^{2} + 231 \, a^{3} d^{2} e^{3} + 35 \,{\left (4 \, b^{3} d e^{4} + 11 \, a b^{2} e^{5}\right )} x^{4} + 5 \,{\left (b^{3} d^{2} e^{3} + 110 \, a b^{2} d e^{4} + 99 \, a^{2} b e^{5}\right )} x^{3} - 3 \,{\left (2 \, b^{3} d^{3} e^{2} - 11 \, a b^{2} d^{2} e^{3} - 264 \, a^{2} b d e^{4} - 77 \, a^{3} e^{5}\right )} x^{2} +{\left (8 \, b^{3} d^{4} e - 44 \, a b^{2} d^{3} e^{2} + 99 \, a^{2} b d^{2} e^{3} + 462 \, a^{3} d e^{4}\right )} x\right )} \sqrt{e x + d}}{1155 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

2/1155*(105*b^3*e^5*x^5 - 16*b^3*d^5 + 88*a*b^2*d^4*e - 198*a^2*b*d^3*e^2 + 231*a^3*d^2*e^3 + 35*(4*b^3*d*e^4
+ 11*a*b^2*e^5)*x^4 + 5*(b^3*d^2*e^3 + 110*a*b^2*d*e^4 + 99*a^2*b*e^5)*x^3 - 3*(2*b^3*d^3*e^2 - 11*a*b^2*d^2*e
^3 - 264*a^2*b*d*e^4 - 77*a^3*e^5)*x^2 + (8*b^3*d^4*e - 44*a*b^2*d^3*e^2 + 99*a^2*b*d^2*e^3 + 462*a^3*d*e^4)*x
)*sqrt(e*x + d)/e^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{\frac{3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1583, size = 528, normalized size = 2.54 \begin{align*} \frac{2}{3465} \,{\left (693 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a^{2} b d e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 99 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} a b^{2} d e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 11 \,{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} b^{3} d e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) + 1155 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} d \mathrm{sgn}\left (b x + a\right ) + 99 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} a^{2} b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 33 \,{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} a b^{2} e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) +{\left (315 \,{\left (x e + d\right )}^{\frac{11}{2}} - 1540 \,{\left (x e + d\right )}^{\frac{9}{2}} d + 2970 \,{\left (x e + d\right )}^{\frac{7}{2}} d^{2} - 2772 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{3} + 1155 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{4}\right )} b^{3} e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) + 231 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

2/3465*(693*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^2*b*d*e^(-1)*sgn(b*x + a) + 99*(15*(x*e + d)^(7/2) - 4
2*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*a*b^2*d*e^(-2)*sgn(b*x + a) + 11*(35*(x*e + d)^(9/2) - 135*(x*e
+ d)^(7/2)*d + 189*(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*b^3*d*e^(-3)*sgn(b*x + a) + 1155*(x*e + d)^(
3/2)*a^3*d*sgn(b*x + a) + 99*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*a^2*b*e^(-1)
*sgn(b*x + a) + 33*(35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*(x*e + d)^(5/2)*d^2 - 105*(x*e + d)^(3/2)
*d^3)*a*b^2*e^(-2)*sgn(b*x + a) + (315*(x*e + d)^(11/2) - 1540*(x*e + d)^(9/2)*d + 2970*(x*e + d)^(7/2)*d^2 -
2772*(x*e + d)^(5/2)*d^3 + 1155*(x*e + d)^(3/2)*d^4)*b^3*e^(-3)*sgn(b*x + a) + 231*(3*(x*e + d)^(5/2) - 5*(x*e
+ d)^(3/2)*d)*a^3*sgn(b*x + a))*e^(-1)