3.228 \(\int x^5 (c (a+b x^2)^2)^{3/2} \, dx\)

Optimal. Leaf size=143 \[ \frac{3 a^2 b c x^8 \sqrt{c \left (a+b x^2\right )^2}}{8 \left (a+b x^2\right )}+\frac{a^3 c x^6 \sqrt{c \left (a+b x^2\right )^2}}{6 \left (a+b x^2\right )}+\frac{b^3 c x^{12} \sqrt{c \left (a+b x^2\right )^2}}{12 \left (a+b x^2\right )}+\frac{3 a b^2 c x^{10} \sqrt{c \left (a+b x^2\right )^2}}{10 \left (a+b x^2\right )} \]

[Out]

(a^3*c*x^6*Sqrt[c*(a + b*x^2)^2])/(6*(a + b*x^2)) + (3*a^2*b*c*x^8*Sqrt[c*(a + b*x^2)^2])/(8*(a + b*x^2)) + (3
*a*b^2*c*x^10*Sqrt[c*(a + b*x^2)^2])/(10*(a + b*x^2)) + (b^3*c*x^12*Sqrt[c*(a + b*x^2)^2])/(12*(a + b*x^2))

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Rubi [A]  time = 0.161853, antiderivative size = 134, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1989, 1111, 645} \[ \frac{c \left (a+b x^2\right )^5 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{12 b^3}-\frac{a c \left (a+b x^2\right )^4 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{5 b^3}+\frac{a^2 c \left (a+b x^2\right )^3 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(a^2*c*(a + b*x^2)^3*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(8*b^3) - (a*c*(a + b*x^2)^4*Sqrt[a^2*c + 2*a*b*c*
x^2 + b^2*c*x^4])/(5*b^3) + (c*(a + b*x^2)^5*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(12*b^3)

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 645

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

Rubi steps

\begin{align*} \int x^5 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int x^5 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a^2 c+2 a b c x+b^2 c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \operatorname{Subst}\left (\int \left (\frac{a^2 \left (a b c+b^2 c x\right )^3}{b^2}-\frac{2 a \left (a b c+b^2 c x\right )^4}{b^3 c}+\frac{\left (a b c+b^2 c x\right )^5}{b^4 c^2}\right ) \, dx,x,x^2\right )}{2 b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac{a^2 c \left (a+b x^2\right )^3 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{8 b^3}-\frac{a c \left (a+b x^2\right )^4 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{5 b^3}+\frac{c \left (a+b x^2\right )^5 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{12 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0215343, size = 63, normalized size = 0.44 \[ \frac{x^6 \left (45 a^2 b x^2+20 a^3+36 a b^2 x^4+10 b^3 x^6\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{120 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^6*(c*(a + b*x^2)^2)^(3/2)*(20*a^3 + 45*a^2*b*x^2 + 36*a*b^2*x^4 + 10*b^3*x^6))/(120*(a + b*x^2)^3)

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Maple [A]  time = 0.004, size = 60, normalized size = 0.4 \begin{align*}{\frac{{x}^{6} \left ( 10\,{b}^{3}{x}^{6}+36\,a{b}^{2}{x}^{4}+45\,{a}^{2}b{x}^{2}+20\,{a}^{3} \right ) }{120\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( c \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*x^6*(10*b^3*x^6+36*a*b^2*x^4+45*a^2*b*x^2+20*a^3)*(c*(b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.41231, size = 166, normalized size = 1.16 \begin{align*} \frac{{\left (10 \, b^{3} c x^{12} + 36 \, a b^{2} c x^{10} + 45 \, a^{2} b c x^{8} + 20 \, a^{3} c x^{6}\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{120 \,{\left (b x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(10*b^3*c*x^12 + 36*a*b^2*c*x^10 + 45*a^2*b*c*x^8 + 20*a^3*c*x^6)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/
(b*x^2 + a)

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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(x**5*(c*(b*x**2+a)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21836, size = 97, normalized size = 0.68 \begin{align*} \frac{1}{120} \,{\left (10 \, b^{3} x^{12} \mathrm{sgn}\left (b x^{2} + a\right ) + 36 \, a b^{2} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + 45 \, a^{2} b x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + 20 \, a^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right )\right )} c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(10*b^3*x^12*sgn(b*x^2 + a) + 36*a*b^2*x^10*sgn(b*x^2 + a) + 45*a^2*b*x^8*sgn(b*x^2 + a) + 20*a^3*x^6*sg
n(b*x^2 + a))*c^(3/2)