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

Optimal. Leaf size=134 \[ \frac{3 a^2 b c x \sqrt{c \left (a+b x^2\right )^2}}{a+b x^2}-\frac{a^3 c \sqrt{c \left (a+b x^2\right )^2}}{x \left (a+b x^2\right )}+\frac{b^3 c x^5 \sqrt{c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )}+\frac{a b^2 c x^3 \sqrt{c \left (a+b x^2\right )^2}}{a+b x^2} \]

[Out]

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

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Rubi [A]  time = 0.0933017, antiderivative size = 178, normalized size of antiderivative = 1.33, 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} \[ \frac{b^3 c x^5 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{5 \left (a+b x^2\right )}+\frac{a b^2 c x^3 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{a+b x^2}+\frac{3 a^2 b c x \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{a+b x^2}-\frac{a^3 c \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x^2)^2)^(3/2)/x^2,x]

[Out]

-((a^3*c*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(x*(a + b*x^2))) + (3*a^2*b*c*x*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2
*c*x^4])/(a + b*x^2) + (a*b^2*c*x^3*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/(a + b*x^2) + (b^3*c*x^5*Sqrt[a^2*c
+ 2*a*b*c*x^2 + b^2*c*x^4])/(5*(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 \frac{\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x^2} \, dx &=\int \frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{x^2} \, dx\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \int \frac{\left (a b c+b^2 c x^2\right )^3}{x^2} \, 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 (3 a^2 b^4 c^3+\frac{a^3 b^3 c^3}{x^2}+3 a b^5 c^3 x^2+b^6 c^3 x^4\right ) \, dx}{b^2 c \left (a b c+b^2 c x^2\right )}\\ &=-\frac{a^3 c \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{x \left (a+b x^2\right )}+\frac{3 a^2 b c x \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{a+b x^2}+\frac{a b^2 c x^3 \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{a+b x^2}+\frac{b^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 )}\\ \end{align*}

Mathematica [A]  time = 0.0237818, size = 62, normalized size = 0.46 \[ \frac{\left (15 a^2 b x^2-5 a^3+5 a b^2 x^4+b^3 x^6\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{5 x \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*(a + b*x^2)^2)^(3/2)*(-5*a^3 + 15*a^2*b*x^2 + 5*a*b^2*x^4 + b^3*x^6))/(5*x*(a + b*x^2)^3)

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Maple [A]  time = 0.005, size = 60, normalized size = 0.5 \begin{align*} -{\frac{-{b}^{3}{x}^{6}-5\,a{b}^{2}{x}^{4}-15\,{a}^{2}b{x}^{2}+5\,{a}^{3}}{5\,x \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((c*(b*x^2+a)^2)^(3/2)/x^2,x)

[Out]

-1/5*(-b^3*x^6-5*a*b^2*x^4-15*a^2*b*x^2+5*a^3)*(c*(b*x^2+a)^2)^(3/2)/x/(b*x^2+a)^3

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Maxima [A]  time = 1.02658, size = 65, normalized size = 0.49 \begin{align*} \frac{b^{3} c^{\frac{3}{2}} x^{6} + 5 \, a b^{2} c^{\frac{3}{2}} x^{4} + 15 \, a^{2} b c^{\frac{3}{2}} x^{2} - 5 \, a^{3} c^{\frac{3}{2}}}{5 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*c^(3/2)*x^6 + 5*a*b^2*c^(3/2)*x^4 + 15*a^2*b*c^(3/2)*x^2 - 5*a^3*c^(3/2))/x

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Fricas [A]  time = 1.41895, size = 151, normalized size = 1.13 \begin{align*} \frac{{\left (b^{3} c x^{6} + 5 \, a b^{2} c x^{4} + 15 \, a^{2} b c x^{2} - 5 \, a^{3} c\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{5 \,{\left (b x^{3} + a x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*c*x^6 + 5*a*b^2*c*x^4 + 15*a^2*b*c*x^2 - 5*a^3*c)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^3 + a*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25013, size = 93, normalized size = 0.69 \begin{align*} \frac{1}{5} \,{\left (b^{3} x^{5} \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, a b^{2} x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, a^{2} b x \mathrm{sgn}\left (b x^{2} + a\right ) - \frac{5 \, a^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{x}\right )} c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*x^5*sgn(b*x^2 + a) + 5*a*b^2*x^3*sgn(b*x^2 + a) + 15*a^2*b*x*sgn(b*x^2 + a) - 5*a^3*sgn(b*x^2 + a)/x)
*c^(3/2)