3.229 \(\int x^4 (c (a+b x^2)^2)^{3/2} \, dx\)

Optimal. Leaf size=143 \[ \frac{3 a^2 b c x^7 \sqrt{c \left (a+b x^2\right )^2}}{7 \left (a+b x^2\right )}+\frac{a^3 c x^5 \sqrt{c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )}+\frac{b^3 c x^{11} \sqrt{c \left (a+b x^2\right )^2}}{11 \left (a+b x^2\right )}+\frac{a b^2 c x^9 \sqrt{c \left (a+b x^2\right )^2}}{3 \left (a+b x^2\right )} \]

[Out]

(a^3*c*x^5*Sqrt[c*(a + b*x^2)^2])/(5*(a + b*x^2)) + (3*a^2*b*c*x^7*Sqrt[c*(a + b*x^2)^2])/(7*(a + b*x^2)) + (a
*b^2*c*x^9*Sqrt[c*(a + b*x^2)^2])/(3*(a + b*x^2)) + (b^3*c*x^11*Sqrt[c*(a + b*x^2)^2])/(11*(a + b*x^2))

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Rubi [A]  time = 0.10539, antiderivative size = 187, normalized size of antiderivative = 1.31, 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, 1112, 270} \[ \frac{b^3 c x^{11} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{11 \left (a+b x^2\right )}+\frac{a b^2 c x^9 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{3 \left (a+b x^2\right )}+\frac{3 a^2 b c x^7 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{7 \left (a+b x^2\right )}+\frac{a^3 c x^5 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(a^3*c*x^5*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(5*(a + b*x^2)) + (3*a^2*b*c*x^7*Sqrt[a^2*c + 2*a*b*c*x^2 +
b^2*c*x^4])/(7*(a + b*x^2)) + (a*b^2*c*x^9*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(3*(a + b*x^2)) + (b^3*c*x^1
1*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(11*(a + b*x^2))

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 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x^4 \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int x^4 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \int x^4 \left (a b c+b^2 c x^2\right )^3 \, dx}{b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \int \left (a^3 b^3 c^3 x^4+3 a^2 b^4 c^3 x^6+3 a b^5 c^3 x^8+b^6 c^3 x^{10}\right ) \, dx}{b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac{a^3 c x^5 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{5 \left (a+b x^2\right )}+\frac{3 a^2 b c x^7 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{7 \left (a+b x^2\right )}+\frac{a b^2 c x^9 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{3 \left (a+b x^2\right )}+\frac{b^3 c x^{11} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{11 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0213691, size = 63, normalized size = 0.44 \[ \frac{x^5 \left (495 a^2 b x^2+231 a^3+385 a b^2 x^4+105 b^3 x^6\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{1155 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(x^5*(c*(a + b*x^2)^2)^(3/2)*(231*a^3 + 495*a^2*b*x^2 + 385*a*b^2*x^4 + 105*b^3*x^6))/(1155*(a + b*x^2)^3)

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Maple [A]  time = 0.003, size = 60, normalized size = 0.4 \begin{align*}{\frac{{x}^{5} \left ( 105\,{b}^{3}{x}^{6}+385\,a{b}^{2}{x}^{4}+495\,{a}^{2}b{x}^{2}+231\,{a}^{3} \right ) }{1155\, \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^4*(c*(b*x^2+a)^2)^(3/2),x)

[Out]

1/1155*x^5*(105*b^3*x^6+385*a*b^2*x^4+495*a^2*b*x^2+231*a^3)*(c*(b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [A]  time = 0.996651, size = 63, normalized size = 0.44 \begin{align*} \frac{1}{11} \, b^{3} c^{\frac{3}{2}} x^{11} + \frac{1}{3} \, a b^{2} c^{\frac{3}{2}} x^{9} + \frac{3}{7} \, a^{2} b c^{\frac{3}{2}} x^{7} + \frac{1}{5} \, a^{3} c^{\frac{3}{2}} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*(b*x^2+a)^2)^(3/2),x, algorithm="maxima")

[Out]

1/11*b^3*c^(3/2)*x^11 + 1/3*a*b^2*c^(3/2)*x^9 + 3/7*a^2*b*c^(3/2)*x^7 + 1/5*a^3*c^(3/2)*x^5

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Fricas [A]  time = 1.40636, size = 171, normalized size = 1.2 \begin{align*} \frac{{\left (105 \, b^{3} c x^{11} + 385 \, a b^{2} c x^{9} + 495 \, a^{2} b c x^{7} + 231 \, a^{3} c x^{5}\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{1155 \,{\left (b x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(c*(b*x^2+a)^2)^(3/2),x, algorithm="fricas")

[Out]

1/1155*(105*b^3*c*x^11 + 385*a*b^2*c*x^9 + 495*a^2*b*c*x^7 + 231*a^3*c*x^5)*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**4*(c*(b*x**2+a)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27508, size = 97, normalized size = 0.68 \begin{align*} \frac{1}{1155} \,{\left (105 \, b^{3} x^{11} \mathrm{sgn}\left (b x^{2} + a\right ) + 385 \, a b^{2} x^{9} \mathrm{sgn}\left (b x^{2} + a\right ) + 495 \, a^{2} b x^{7} \mathrm{sgn}\left (b x^{2} + a\right ) + 231 \, a^{3} x^{5} \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^4*(c*(b*x^2+a)^2)^(3/2),x, algorithm="giac")

[Out]

1/1155*(105*b^3*x^11*sgn(b*x^2 + a) + 385*a*b^2*x^9*sgn(b*x^2 + a) + 495*a^2*b*x^7*sgn(b*x^2 + a) + 231*a^3*x^
5*sgn(b*x^2 + a))*c^(3/2)