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

Optimal. Leaf size=208 \[ \frac{315 a^3 b c x \sqrt{c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{315 a^{5/2} \sqrt{b} c \sqrt{c \left (a+b x^2\right )^3} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 \left (\frac{b x^2}{a}+1\right )^{3/2}}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3} \]

[Out]

(105*a^2*b*c*x*Sqrt[c*(a + b*x^2)^3])/64 + (315*a^3*b*c*x*Sqrt[c*(a + b*x^2)^3])/(128*(a + b*x^2)) + (21*a*b*c
*x*(a + b*x^2)*Sqrt[c*(a + b*x^2)^3])/16 + (9*b*c*x*(a + b*x^2)^2*Sqrt[c*(a + b*x^2)^3])/8 - (c*(a + b*x^2)^3*
Sqrt[c*(a + b*x^2)^3])/x + (315*a^(5/2)*Sqrt[b]*c*Sqrt[c*(a + b*x^2)^3]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(128*(1
+ (b*x^2)/a)^(3/2))

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Rubi [A]  time = 0.18896, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {6720, 277, 195, 217, 206} \[ \frac{315 a^3 b c x \sqrt{c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{315 a^4 \sqrt{b} c \sqrt{c \left (a+b x^2\right )^3} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 \left (a+b x^2\right )^{3/2}}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x^2)^3)^(3/2)/x^2,x]

[Out]

(105*a^2*b*c*x*Sqrt[c*(a + b*x^2)^3])/64 + (315*a^3*b*c*x*Sqrt[c*(a + b*x^2)^3])/(128*(a + b*x^2)) + (21*a*b*c
*x*(a + b*x^2)*Sqrt[c*(a + b*x^2)^3])/16 + (9*b*c*x*(a + b*x^2)^2*Sqrt[c*(a + b*x^2)^3])/8 - (c*(a + b*x^2)^3*
Sqrt[c*(a + b*x^2)^3])/x + (315*a^4*Sqrt[b]*c*Sqrt[c*(a + b*x^2)^3]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(128
*(a + b*x^2)^(3/2))

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^2} \, dx &=\frac{\left (c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \frac{\left (a+b x^2\right )^{9/2}}{x^2} \, dx}{\left (a+b x^2\right )^{3/2}}\\ &=-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (9 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \left (a+b x^2\right )^{7/2} \, dx}{\left (a+b x^2\right )^{3/2}}\\ &=\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (63 a b c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \left (a+b x^2\right )^{5/2} \, dx}{8 \left (a+b x^2\right )^{3/2}}\\ &=\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (105 a^2 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{16 \left (a+b x^2\right )^{3/2}}\\ &=\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (315 a^3 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \sqrt{a+b x^2} \, dx}{64 \left (a+b x^2\right )^{3/2}}\\ &=\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{315 a^3 b c x \sqrt{c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (315 a^4 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{128 \left (a+b x^2\right )^{3/2}}\\ &=\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{315 a^3 b c x \sqrt{c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{\left (315 a^4 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{128 \left (a+b x^2\right )^{3/2}}\\ &=\frac{105}{64} a^2 b c x \sqrt{c \left (a+b x^2\right )^3}+\frac{315 a^3 b c x \sqrt{c \left (a+b x^2\right )^3}}{128 \left (a+b x^2\right )}+\frac{21}{16} a b c x \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{8} b c x \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{x}+\frac{315 a^4 \sqrt{b} c \sqrt{c \left (a+b x^2\right )^3} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 \left (a+b x^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0159161, size = 65, normalized size = 0.31 \[ -\frac{a^4 \left (c \left (a+b x^2\right )^3\right )^{3/2} \, _2F_1\left (-\frac{9}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \left (a+b x^2\right )^4 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^4*(c*(a + b*x^2)^3)^(3/2)*Hypergeometric2F1[-9/2, -1/2, 1/2, -((b*x^2)/a)])/(x*(a + b*x^2)^4*Sqrt[1 + (b*
x^2)/a]))

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Maple [A]  time = 0.008, size = 215, normalized size = 1. \begin{align*}{\frac{1}{128\,c \left ( b{x}^{2}+a \right ) ^{3}x} \left ( c \left ( b{x}^{2}+a \right ) ^{3} \right ) ^{{\frac{3}{2}}} \left ( 16\, \left ( bc{x}^{2}+ac \right ) ^{5/2}\sqrt{bc}{x}^{4}{b}^{2}+56\, \left ( bc{x}^{2}+ac \right ) ^{5/2}\sqrt{bc}{x}^{2}ab+210\, \left ( bc{x}^{2}+ac \right ) ^{3/2}\sqrt{bc}{x}^{2}{a}^{2}bc+315\,\sqrt{bc{x}^{2}+ac}\sqrt{bc}{x}^{2}{a}^{3}b{c}^{2}+315\,\ln \left ({\frac{bcx+\sqrt{bc{x}^{2}+ac}\sqrt{bc}}{\sqrt{bc}}} \right ) x{a}^{4}b{c}^{3}-128\, \left ( bc{x}^{2}+ac \right ) ^{5/2}\sqrt{bc}{a}^{2} \right ) \left ( c \left ( b{x}^{2}+a \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(c*(b*x^2+a)^3)^(3/2)*(16*(b*c*x^2+a*c)^(5/2)*(b*c)^(1/2)*x^4*b^2+56*(b*c*x^2+a*c)^(5/2)*(b*c)^(1/2)*x^2
*a*b+210*(b*c*x^2+a*c)^(3/2)*(b*c)^(1/2)*x^2*a^2*b*c+315*(b*c*x^2+a*c)^(1/2)*(b*c)^(1/2)*x^2*a^3*b*c^2+315*ln(
(b*c*x+(b*c*x^2+a*c)^(1/2)*(b*c)^(1/2))/(b*c)^(1/2))*x*a^4*b*c^3-128*(b*c*x^2+a*c)^(5/2)*(b*c)^(1/2)*a^2)/(b*x
^2+a)^3/(c*(b*x^2+a))^(3/2)/c/(b*c)^(1/2)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74705, size = 873, normalized size = 4.2 \begin{align*} \left [\frac{315 \,{\left (a^{4} b c x^{3} + a^{5} c x\right )} \sqrt{b c} \log \left (-\frac{2 \, b^{2} c x^{4} + 3 \, a b c x^{2} + a^{2} c + 2 \, \sqrt{b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt{b c} x}{b x^{2} + a}\right ) + 2 \,{\left (16 \, b^{4} c x^{8} + 88 \, a b^{3} c x^{6} + 210 \, a^{2} b^{2} c x^{4} + 325 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt{b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{256 \,{\left (b x^{3} + a x\right )}}, -\frac{315 \,{\left (a^{4} b c x^{3} + a^{5} c x\right )} \sqrt{-b c} \arctan \left (\frac{\sqrt{b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt{-b c} x}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) -{\left (16 \, b^{4} c x^{8} + 88 \, a b^{3} c x^{6} + 210 \, a^{2} b^{2} c x^{4} + 325 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt{b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{128 \,{\left (b x^{3} + a x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/256*(315*(a^4*b*c*x^3 + a^5*c*x)*sqrt(b*c)*log(-(2*b^2*c*x^4 + 3*a*b*c*x^2 + a^2*c + 2*sqrt(b^3*c*x^6 + 3*a
*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c)*sqrt(b*c)*x)/(b*x^2 + a)) + 2*(16*b^4*c*x^8 + 88*a*b^3*c*x^6 + 210*a^2*b^2
*c*x^4 + 325*a^3*b*c*x^2 - 128*a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c))/(b*x^3 + a*x),
-1/128*(315*(a^4*b*c*x^3 + a^5*c*x)*sqrt(-b*c)*arctan(sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*c)*
sqrt(-b*c)*x/(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)) - (16*b^4*c*x^8 + 88*a*b^3*c*x^6 + 210*a^2*b^2*c*x^4 + 325*a^3
*b*c*x^2 - 128*a^4*c)*sqrt(b^3*c*x^6 + 3*a*b^2*c*x^4 + 3*a^2*b*c*x^2 + a^3*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 )^{3}\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)**3)**(3/2)/x**2,x)

[Out]

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

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Giac [A]  time = 1.33512, size = 250, normalized size = 1.2 \begin{align*} \frac{1}{256} \,{\left (\frac{512 \, \sqrt{b c} a^{5} c \mathrm{sgn}\left (b x^{2} + a\right )}{{\left (\sqrt{b c} x - \sqrt{b c x^{2} + a c}\right )}^{2} - a c} - 315 \, \sqrt{b c} a^{4} \log \left ({\left (\sqrt{b c} x - \sqrt{b c x^{2} + a c}\right )}^{2}\right ) \mathrm{sgn}\left (b x^{2} + a\right ) + 2 \,{\left (325 \, a^{3} b \mathrm{sgn}\left (b x^{2} + a\right ) + 2 \,{\left (105 \, a^{2} b^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 4 \,{\left (2 \, b^{4} x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 11 \, a b^{3} \mathrm{sgn}\left (b x^{2} + a\right )\right )} x^{2}\right )} x^{2}\right )} \sqrt{b c x^{2} + a c} x\right )} c \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(512*sqrt(b*c)*a^5*c*sgn(b*x^2 + a)/((sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^2 - a*c) - 315*sqrt(b*c)*a^4*lo
g((sqrt(b*c)*x - sqrt(b*c*x^2 + a*c))^2)*sgn(b*x^2 + a) + 2*(325*a^3*b*sgn(b*x^2 + a) + 2*(105*a^2*b^2*sgn(b*x
^2 + a) + 4*(2*b^4*x^2*sgn(b*x^2 + a) + 11*a*b^3*sgn(b*x^2 + a))*x^2)*x^2)*sqrt(b*c*x^2 + a*c)*x)*c