3.17.85 \(\int \sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [1685]

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

[Out]

-2/3*(-a*e+b*d)^3*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+6/5*b*(-a*e+b*d)^2*(e*x+d)^(5/2)*((b*x+a)^2)^(1/
2)/e^4/(b*x+a)-6/7*b^2*(-a*e+b*d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/9*b^3*(e*x+d)^(9/2)*((b*x+a)^2
)^(1/2)/e^4/(b*x+a)

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Rubi [A]
time = 0.05, 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 = {660, 45} \begin {gather*} -\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^4 (a+b x)}+\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}{5 e^4 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}{3 e^4 (a+b x)}+\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 660

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

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Mathematica [A]
time = 0.07, size = 120, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} \left (105 a^3 e^3+63 a^2 b e^2 (-2 d+3 e x)+9 a b^2 e \left (8 d^2-12 d e x+15 e^2 x^2\right )+b^3 \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )}{315 e^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(3/2)*(105*a^3*e^3 + 63*a^2*b*e^2*(-2*d + 3*e*x) + 9*a*b^2*e*(8*d^2 - 12*d*e*x
+ 15*e^2*x^2) + b^3*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 35*e^3*x^3)))/(315*e^4*(a + b*x))

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Maple [A]
time = 0.64, size = 132, normalized size = 0.63

method	result	size
gosper	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 b^{3} x^{3} e^{3}+135 a \,b^{2} e^{3} x^{2}-30 b^{3} d \,e^{2} x^{2}+189 a^{2} b \,e^{3} x -108 a \,b^{2} d \,e^{2} x +24 b^{3} d^{2} e x +105 e^{3} a^{3}-126 a^{2} b d \,e^{2}+72 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 e^{4} \left (b x +a \right )^{3}}\)	\(132\)
default	\(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 b^{3} x^{3} e^{3}+135 a \,b^{2} e^{3} x^{2}-30 b^{3} d \,e^{2} x^{2}+189 a^{2} b \,e^{3} x -108 a \,b^{2} d \,e^{2} x +24 b^{3} d^{2} e x +105 e^{3} a^{3}-126 a^{2} b d \,e^{2}+72 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 e^{4} \left (b x +a \right )^{3}}\)	\(132\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (35 b^{3} x^{4} e^{4}+135 a \,b^{2} e^{4} x^{3}+5 b^{3} d \,e^{3} x^{3}+189 a^{2} b \,e^{4} x^{2}+27 a \,b^{2} d \,e^{3} x^{2}-6 b^{3} d^{2} e^{2} x^{2}+105 a^{3} e^{4} x +63 a^{2} b d \,e^{3} x -36 a \,b^{2} d^{2} e^{2} x +8 b^{3} d^{3} e x +105 a^{3} d \,e^{3}-126 a^{2} b \,d^{2} e^{2}+72 a \,b^{2} d^{3} e -16 b^{3} d^{4}\right ) \sqrt {e x +d}}{315 \left (b x +a \right ) e^{4}}\)	\(186\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/315*(e*x+d)^(3/2)*(35*b^3*e^3*x^3+135*a*b^2*e^3*x^2-30*b^3*d*e^2*x^2+189*a^2*b*e^3*x-108*a*b^2*d*e^2*x+24*b^
3*d^2*e*x+105*a^3*e^3-126*a^2*b*d*e^2+72*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 = 0.29, size = 155, normalized size = 0.75 \begin {gather*} \frac {2}{315} \, {\left (35 \, b^{3} x^{4} e^{4} - 16 \, b^{3} d^{4} + 72 \, a b^{2} d^{3} e - 126 \, a^{2} b d^{2} e^{2} + 105 \, a^{3} d e^{3} + 5 \, {\left (b^{3} d e^{3} + 27 \, a b^{2} e^{4}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{2} e^{2} - 9 \, a b^{2} d e^{3} - 63 \, a^{2} b e^{4}\right )} x^{2} + {\left (8 \, b^{3} d^{3} e - 36 \, a b^{2} d^{2} e^{2} + 63 \, a^{2} b d e^{3} + 105 \, a^{3} e^{4}\right )} x\right )} \sqrt {x e + d} e^{\left (-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)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*b^3*x^4*e^4 - 16*b^3*d^4 + 72*a*b^2*d^3*e - 126*a^2*b*d^2*e^2 + 105*a^3*d*e^3 + 5*(b^3*d*e^3 + 27*a*
b^2*e^4)*x^3 - 3*(2*b^3*d^2*e^2 - 9*a*b^2*d*e^3 - 63*a^2*b*e^4)*x^2 + (8*b^3*d^3*e - 36*a*b^2*d^2*e^2 + 63*a^2
*b*d*e^3 + 105*a^3*e^4)*x)*sqrt(x*e + d)*e^(-4)

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Fricas [A]
time = 2.24, size = 152, normalized size = 0.73 \begin {gather*} -\frac {2}{315} \, {\left (16 \, b^{3} d^{4} - {\left (35 \, b^{3} x^{4} + 135 \, a b^{2} x^{3} + 189 \, a^{2} b x^{2} + 105 \, a^{3} x\right )} e^{4} - {\left (5 \, b^{3} d x^{3} + 27 \, a b^{2} d x^{2} + 63 \, a^{2} b d x + 105 \, a^{3} d\right )} e^{3} + 6 \, {\left (b^{3} d^{2} x^{2} + 6 \, a b^{2} d^{2} x + 21 \, a^{2} b d^{2}\right )} e^{2} - 8 \, {\left (b^{3} d^{3} x + 9 \, a b^{2} d^{3}\right )} e\right )} \sqrt {x e + d} e^{\left (-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)^(1/2),x, algorithm="fricas")

[Out]

-2/315*(16*b^3*d^4 - (35*b^3*x^4 + 135*a*b^2*x^3 + 189*a^2*b*x^2 + 105*a^3*x)*e^4 - (5*b^3*d*x^3 + 27*a*b^2*d*
x^2 + 63*a^2*b*d*x + 105*a^3*d)*e^3 + 6*(b^3*d^2*x^2 + 6*a*b^2*d^2*x + 21*a^2*b*d^2)*e^2 - 8*(b^3*d^3*x + 9*a*
b^2*d^3)*e)*sqrt(x*e + d)*e^(-4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{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)**(1/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (151) = 302\).
time = 1.01, size = 387, normalized size = 1.86 \begin {gather*} \frac {2}{315} \, {\left (315 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a^{2} b d e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 63 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} a b^{2} d e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 9 \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} b^{3} d e^{\left (-3\right )} \mathrm {sgn}\left (b x + a\right ) + 63 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} a^{2} b e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 27 \, {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} a b^{2} e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + {\left (35 \, {\left (x e + d\right )}^{\frac {9}{2}} - 180 \, {\left (x e + d\right )}^{\frac {7}{2}} d + 378 \, {\left (x e + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {x e + d} d^{4}\right )} b^{3} e^{\left (-3\right )} \mathrm {sgn}\left (b x + a\right ) + 315 \, \sqrt {x e + d} a^{3} d \mathrm {sgn}\left (b x + a\right ) + 105 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\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)^(1/2),x, algorithm="giac")

[Out]

2/315*(315*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^2*b*d*e^(-1)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e
+ d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a*b^2*d*e^(-2)*sgn(b*x + a) + 9*(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*e^(-3)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e
+ d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a^2*b*e^(-1)*sgn(b*x + a) + 27*(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*e^(-2)*sgn(b*x + a) + (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)*b^3*e^(-3)*sgn(b*x +
a) + 315*sqrt(x*e + d)*a^3*d*sgn(b*x + a) + 105*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^3*sgn(b*x + a))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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