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

[Out]

-2/7*(-a*e+b*d)^3*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/3*b*(-a*e+b*d)^2*(e*x+d)^(9/2)*((b*x+a)^2)^(1/
2)/e^4/(b*x+a)-6/11*b^2*(-a*e+b*d)*(e*x+d)^(11/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/13*b^3*(e*x+d)^(13/2)*((b*x+
a)^2)^(1/2)/e^4/(b*x+a)

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Rubi [A]  time = 0.07, antiderivative size = 208, 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} \[ \frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{13/2}}{13 e^4 (a+b x)}-\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{11/2} (b d-a e)}{11 e^4 (a+b x)}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2} (b d-a e)^2}{3 e^4 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)^3}{7 e^4 (a+b x)} \]

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 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 (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*}

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Mathematica [A]  time = 0.09, size = 120, normalized size = 0.58 \[ \frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} \left (429 a^3 e^3+143 a^2 b e^2 (7 e x-2 d)+13 a b^2 e \left (8 d^2-28 d e x+63 e^2 x^2\right )+b^3 \left (-16 d^3+56 d^2 e x-126 d e^2 x^2+231 e^3 x^3\right )\right )}{3003 e^4 (a+b x)} \]

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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fricas [A]  time = 1.03, size = 268, normalized size = 1.29 \[ \frac {2 \, {\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \, {\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \, {\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} + {\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{3003 \, e^{4}} \]

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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giac [B]  time = 0.27, size = 1004, normalized size = 4.83 \[ \text {result too large to display} \]

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]

2/15015*(15015*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^2*b*d^3*e^(-1)*sgn(b*x + a) + 3003*(3*(x*e + d)^(5/2) -
10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a*b^2*d^3*e^(-2)*sgn(b*x + a) + 429*(5*(x*e + d)^(7/2) - 21*(x*e
+ d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*b^3*d^3*e^(-3)*sgn(b*x + a) + 9009*(3*(x*e + d)
^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a^2*b*d^2*e^(-1)*sgn(b*x + a) + 3861*(5*(x*e + d)^(7/2)
- 21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*a*b^2*d^2*e^(-2)*sgn(b*x + a) + 143*(3
5*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(x*e + d)^(5/2)*d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e +
d)*d^4)*b^3*d^2*e^(-3)*sgn(b*x + a) + 15015*sqrt(x*e + d)*a^3*d^3*sgn(b*x + a) + 15015*((x*e + d)^(3/2) - 3*s
qrt(x*e + d)*d)*a^3*d^2*sgn(b*x + a) + 3861*(5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2
- 35*sqrt(x*e + d)*d^3)*a^2*b*d*e^(-1)*sgn(b*x + a) + 429*(35*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(
x*e + d)^(5/2)*d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e + d)*d^4)*a*b^2*d*e^(-2)*sgn(b*x + a) + 65*(63*(x*
e + d)^(11/2) - 385*(x*e + d)^(9/2)*d + 990*(x*e + d)^(7/2)*d^2 - 1386*(x*e + d)^(5/2)*d^3 + 1155*(x*e + d)^(3
/2)*d^4 - 693*sqrt(x*e + d)*d^5)*b^3*d*e^(-3)*sgn(b*x + a) + 3003*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d +
15*sqrt(x*e + d)*d^2)*a^3*d*sgn(b*x + a) + 143*(35*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(x*e + d)^(5/
2)*d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e + d)*d^4)*a^2*b*e^(-1)*sgn(b*x + a) + 65*(63*(x*e + d)^(11/2)
- 385*(x*e + d)^(9/2)*d + 990*(x*e + d)^(7/2)*d^2 - 1386*(x*e + d)^(5/2)*d^3 + 1155*(x*e + d)^(3/2)*d^4 - 693*
sqrt(x*e + d)*d^5)*a*b^2*e^(-2)*sgn(b*x + a) + 5*(231*(x*e + d)^(13/2) - 1638*(x*e + d)^(11/2)*d + 5005*(x*e +
d)^(9/2)*d^2 - 8580*(x*e + d)^(7/2)*d^3 + 9009*(x*e + d)^(5/2)*d^4 - 6006*(x*e + d)^(3/2)*d^5 + 3003*sqrt(x*e
+ d)*d^6)*b^3*e^(-3)*sgn(b*x + a) + 429*(5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 -
35*sqrt(x*e + d)*d^3)*a^3*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.05, size = 132, normalized size = 0.63 \[ \frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (231 b^{3} e^{3} x^{3}+819 a \,b^{2} e^{3} x^{2}-126 b^{3} d \,e^{2} x^{2}+1001 a^{2} b \,e^{3} x -364 a \,b^{2} d \,e^{2} x +56 b^{3} d^{2} e x +429 a^{3} e^{3}-286 a^{2} b d \,e^{2}+104 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3003 \left (b x +a \right )^{3} e^{4}} \]

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 = 1.24, size = 268, normalized size = 1.29 \[ \frac {2 \, {\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \, {\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \, {\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} + {\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{3003 \, e^{4}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (d+e\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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