3.1683 \(\int (d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=208 \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{13/2}}{13 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2} (b d-a e)}{11 e^4 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (b d-a e)^2}{3 e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)^3}{7 e^4 (a+b x)} \]

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^4*(a + b*x)) + (2*b*(b*d - a*e)^2*(d + e
*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^4*(a + b*x)) - (6*b^2*(b*d - a*e)*(d + e*x)^(11/2)*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(11*e^4*(a + b*x)) + (2*b^3*(d + e*x)^(13/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(13*e^4*(a + b*x
))

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Rubi [A]  time = 0.07289, 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} \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{13/2}}{13 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{11/2} (b d-a e)}{11 e^4 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (b d-a e)^2}{3 e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)^3}{7 e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^3*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^4*(a + b*x)) + (2*b*(b*d - a*e)^2*(d + e
*x)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^4*(a + b*x)) - (6*b^2*(b*d - a*e)*(d + e*x)^(11/2)*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(11*e^4*(a + b*x)) + (2*b^3*(d + e*x)^(13/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(13*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)^{5/2} \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)^{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 (d+e x)^{5/2}}{e^3}+\frac{3 b^4 (b d-a e)^2 (d+e x)^{7/2}}{e^3}-\frac{3 b^5 (b d-a e) (d+e x)^{9/2}}{e^3}+\frac{b^6 (d+e x)^{11/2}}{e^3}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{2 (b d-a e)^3 (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 (b d-a e)^2 (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)}-\frac{6 b^2 (b d-a e) (d+e x)^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}{11 e^4 (a+b x)}+\frac{2 b^3 (d+e x)^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}}{13 e^4 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0920237, size = 120, normalized size = 0.58 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{7/2} \left (143 a^2 b e^2 (7 e x-2 d)+429 a^3 e^3+13 a b^2 e \left (8 d^2-28 d e x+63 e^2 x^2\right )+b^3 \left (56 d^2 e x-16 d^3-126 d e^2 x^2+231 e^3 x^3\right )\right )}{3003 e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(429*a^3*e^3 + 143*a^2*b*e^2*(-2*d + 7*e*x) + 13*a*b^2*e*(8*d^2 - 28*d*e*
x + 63*e^2*x^2) + b^3*(-16*d^3 + 56*d^2*e*x - 126*d*e^2*x^2 + 231*e^3*x^3)))/(3003*e^4*(a + b*x))

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Maple [A]  time = 0.154, size = 132, normalized size = 0.6 \begin{align*}{\frac{462\,{x}^{3}{b}^{3}{e}^{3}+1638\,{x}^{2}a{b}^{2}{e}^{3}-252\,{x}^{2}{b}^{3}d{e}^{2}+2002\,x{a}^{2}b{e}^{3}-728\,xa{b}^{2}d{e}^{2}+112\,x{b}^{3}{d}^{2}e+858\,{a}^{3}{e}^{3}-572\,d{e}^{2}{a}^{2}b+208\,a{b}^{2}{d}^{2}e-32\,{b}^{3}{d}^{3}}{3003\,{e}^{4} \left ( bx+a \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{7}{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)^(5/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

2/3003*(e*x+d)^(7/2)*(231*b^3*e^3*x^3+819*a*b^2*e^3*x^2-126*b^3*d*e^2*x^2+1001*a^2*b*e^3*x-364*a*b^2*d*e^2*x+5
6*b^3*d^2*e*x+429*a^3*e^3-286*a^2*b*d*e^2+104*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 = 1.09352, size = 362, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \,{\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \,{\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} +{\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \,{\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} +{\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt{e x + d}}{3003 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

2/3003*(231*b^3*e^6*x^6 - 16*b^3*d^6 + 104*a*b^2*d^5*e - 286*a^2*b*d^4*e^2 + 429*a^3*d^3*e^3 + 63*(9*b^3*d*e^5
+ 13*a*b^2*e^6)*x^5 + 7*(53*b^3*d^2*e^4 + 299*a*b^2*d*e^5 + 143*a^2*b*e^6)*x^4 + (5*b^3*d^3*e^3 + 1469*a*b^2*
d^2*e^4 + 2717*a^2*b*d*e^5 + 429*a^3*e^6)*x^3 - 3*(2*b^3*d^4*e^2 - 13*a*b^2*d^3*e^3 - 715*a^2*b*d^2*e^4 - 429*
a^3*d*e^5)*x^2 + (8*b^3*d^5*e - 52*a*b^2*d^4*e^2 + 143*a^2*b*d^3*e^3 + 1287*a^3*d^2*e^4)*x)*sqrt(e*x + d)/e^4

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Fricas [A]  time = 1.66259, size = 595, normalized size = 2.86 \begin{align*} \frac{2 \,{\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \,{\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \,{\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} +{\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \,{\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} +{\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt{e x + d}}{3003 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

2/3003*(231*b^3*e^6*x^6 - 16*b^3*d^6 + 104*a*b^2*d^5*e - 286*a^2*b*d^4*e^2 + 429*a^3*d^3*e^3 + 63*(9*b^3*d*e^5
+ 13*a*b^2*e^6)*x^5 + 7*(53*b^3*d^2*e^4 + 299*a*b^2*d*e^5 + 143*a^2*b*e^6)*x^4 + (5*b^3*d^3*e^3 + 1469*a*b^2*
d^2*e^4 + 2717*a^2*b*d*e^5 + 429*a^3*e^6)*x^3 - 3*(2*b^3*d^4*e^2 - 13*a*b^2*d^3*e^3 - 715*a^2*b*d^2*e^4 - 429*
a^3*d*e^5)*x^2 + (8*b^3*d^5*e - 52*a*b^2*d^4*e^2 + 143*a^2*b*d^3*e^3 + 1287*a^3*d^2*e^4)*x)*sqrt(e*x + d)/e^4

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.22581, size = 910, normalized size = 4.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

2/45045*(9009*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^2*b*d^2*e^(-1)*sgn(b*x + a) + 1287*(15*(x*e + d)^(7/
2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*a*b^2*d^2*e^(-2)*sgn(b*x + a) + 143*(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^2*e^(-3)*sgn(b*x + a) + 15015
*(x*e + d)^(3/2)*a^3*d^2*sgn(b*x + a) + 2574*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d
^2)*a^2*b*d*e^(-1)*sgn(b*x + a) + 858*(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*d*e^(-2)*sgn(b*x + a) + 26*(315*(x*e + d)^(11/2) - 1540*(x*e + d)^(9/2)*d + 297
0*(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*d*e^(-3)*sgn(b*x + a) + 6006*
(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^3*d*sgn(b*x + a) + 429*(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^2*b*e^(-1)*sgn(b*x + a) + 39*(315*(x*e + d)^(11/2) - 1
540*(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)*a*b^2*
e^(-2)*sgn(b*x + a) + 5*(693*(x*e + d)^(13/2) - 4095*(x*e + d)^(11/2)*d + 10010*(x*e + d)^(9/2)*d^2 - 12870*(x
*e + d)^(7/2)*d^3 + 9009*(x*e + d)^(5/2)*d^4 - 3003*(x*e + d)^(3/2)*d^5)*b^3*e^(-3)*sgn(b*x + a) + 429*(15*(x*
e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*a^3*sgn(b*x + a))*e^(-1)