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

Optimal. Leaf size=202 \[ \frac{9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}-\frac{9 a^2 b c \sqrt{c \left (a+b x^2\right )^3} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{2 \left (\frac{b x^2}{a}+1\right )^{3/2}}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{2 x^2}+\frac{9}{14} b c \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3} \]

[Out]

(3*a^2*b*c*Sqrt[c*(a + b*x^2)^3])/2 + (9*a^3*b*c*Sqrt[c*(a + b*x^2)^3])/(2*(a + b*x^2)) + (9*a*b*c*(a + b*x^2)
*Sqrt[c*(a + b*x^2)^3])/10 + (9*b*c*(a + b*x^2)^2*Sqrt[c*(a + b*x^2)^3])/14 - (c*(a + b*x^2)^3*Sqrt[c*(a + b*x
^2)^3])/(2*x^2) - (9*a^2*b*c*Sqrt[c*(a + b*x^2)^3]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(2*(1 + (b*x^2)/a)^(3/2))

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Rubi [A]  time = 0.215951, antiderivative size = 204, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {6720, 266, 47, 50, 63, 208} \[ \frac{9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}-\frac{9 a^{7/2} b c \sqrt{c \left (a+b x^2\right )^3} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 \left (a+b x^2\right )^{3/2}}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}-\frac{c \left (a+b x^2\right )^3 \sqrt{c \left (a+b x^2\right )^3}}{2 x^2}+\frac{9}{14} b c \left (a+b x^2\right )^2 \sqrt{c \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x^2)^3)^(3/2)/x^3,x]

[Out]

(3*a^2*b*c*Sqrt[c*(a + b*x^2)^3])/2 + (9*a^3*b*c*Sqrt[c*(a + b*x^2)^3])/(2*(a + b*x^2)) + (9*a*b*c*(a + b*x^2)
*Sqrt[c*(a + b*x^2)^3])/10 + (9*b*c*(a + b*x^2)^2*Sqrt[c*(a + b*x^2)^3])/14 - (c*(a + b*x^2)^3*Sqrt[c*(a + b*x
^2)^3])/(2*x^2) - (9*a^(7/2)*b*c*Sqrt[c*(a + b*x^2)^3]*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*(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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\left (c \left (a+b x^2\right )^3\right )^{3/2}}{x^3} \, 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^3} \, dx}{\left (a+b x^2\right )^{3/2}}\\ &=\frac{\left (c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{9/2}}{x^2} \, dx,x,x^2\right )}{2 \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}}{2 x^2}+\frac{\left (9 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{7/2}}{x} \, dx,x,x^2\right )}{4 \left (a+b x^2\right )^{3/2}}\\ &=\frac{9}{14} b c \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}}{2 x^2}+\frac{\left (9 a b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,x^2\right )}{4 \left (a+b x^2\right )^{3/2}}\\ &=\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{14} b c \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}}{2 x^2}+\frac{\left (9 a^2 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{4 \left (a+b x^2\right )^{3/2}}\\ &=\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{14} b c \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}}{2 x^2}+\frac{\left (9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )}{4 \left (a+b x^2\right )^{3/2}}\\ &=\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}+\frac{9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{14} b c \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}}{2 x^2}+\frac{\left (9 a^4 b c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 \left (a+b x^2\right )^{3/2}}\\ &=\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}+\frac{9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{14} b c \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}}{2 x^2}+\frac{\left (9 a^4 c \sqrt{c \left (a+b x^2\right )^3}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 \left (a+b x^2\right )^{3/2}}\\ &=\frac{3}{2} a^2 b c \sqrt{c \left (a+b x^2\right )^3}+\frac{9 a^3 b c \sqrt{c \left (a+b x^2\right )^3}}{2 \left (a+b x^2\right )}+\frac{9}{10} a b c \left (a+b x^2\right ) \sqrt{c \left (a+b x^2\right )^3}+\frac{9}{14} b c \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}}{2 x^2}-\frac{9 a^{7/2} b c \sqrt{c \left (a+b x^2\right )^3} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 \left (a+b x^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0187556, size = 48, normalized size = 0.24 \[ \frac{b \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^3\right )^{3/2} \, _2F_1\left (2,\frac{11}{2};\frac{13}{2};\frac{b x^2}{a}+1\right )}{11 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a + b*x^2)*(c*(a + b*x^2)^3)^(3/2)*Hypergeometric2F1[2, 11/2, 13/2, 1 + (b*x^2)/a])/(11*a^2)

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Maple [A]  time = 0.01, size = 238, normalized size = 1.2 \begin{align*}{\frac{1}{70\,c \left ( b{x}^{2}+a \right ) ^{3}{x}^{2}} \left ( c \left ( b{x}^{2}+a \right ) ^{3} \right ) ^{{\frac{3}{2}}} \left ( 10\,\sqrt{ac} \left ( bc{x}^{2}+ac \right ) ^{5/2}{x}^{4}{b}^{2}-315\,\ln \left ( 2\,{\frac{\sqrt{ac}\sqrt{bc{x}^{2}+ac}+ac}{x}} \right ){x}^{2}{a}^{4}b{c}^{3}-4\,\sqrt{ac} \left ( bc{x}^{2}+ac \right ) ^{5/2}{x}^{2}ab+105\, \left ( bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac}{x}^{2}{a}^{2}bc+315\,\sqrt{bc{x}^{2}+ac}\sqrt{ac}{x}^{2}{a}^{3}b{c}^{2}+42\,ab \left ( c \left ( b{x}^{2}+a \right ) \right ) ^{5/2}\sqrt{ac}{x}^{2}-35\,\sqrt{ac} \left ( bc{x}^{2}+ac \right ) ^{5/2}{a}^{2} \right ) \left ( c \left ( b{x}^{2}+a \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(c*(b*x^2+a)^3)^(3/2)*(10*(a*c)^(1/2)*(b*c*x^2+a*c)^(5/2)*x^4*b^2-315*ln(2*((a*c)^(1/2)*(b*c*x^2+a*c)^(1/
2)+a*c)/x)*x^2*a^4*b*c^3-4*(a*c)^(1/2)*(b*c*x^2+a*c)^(5/2)*x^2*a*b+105*(b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^2*a^2
*b*c+315*(b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^2*a^3*b*c^2+42*a*b*(c*(b*x^2+a))^(5/2)*(a*c)^(1/2)*x^2-35*(a*c)^(1/
2)*(b*c*x^2+a*c)^(5/2)*a^2)/(b*x^2+a)^3/(c*(b*x^2+a))^(3/2)/c/(a*c)^(1/2)/x^2

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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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67576, size = 890, normalized size = 4.41 \begin{align*} \left [\frac{315 \,{\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt{a c} \log \left (-\frac{b^{2} c x^{4} + 3 \, a b c x^{2} + 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{a c}}{b x^{4} + a x^{2}}\right ) + 2 \,{\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, 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}}{140 \,{\left (b x^{4} + a x^{2}\right )}}, \frac{315 \,{\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt{-a 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{-a c}}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) +{\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, 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}}{70 \,{\left (b x^{4} + a x^{2}\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^3,x, algorithm="fricas")

[Out]

[1/140*(315*(a^3*b^2*c*x^4 + a^4*b*c*x^2)*sqrt(a*c)*log(-(b^2*c*x^4 + 3*a*b*c*x^2 + 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(a*c))/(b*x^4 + a*x^2)) + 2*(10*b^4*c*x^8 + 58*a*b^3*c*x^6 + 156
*a^2*b^2*c*x^4 + 388*a^3*b*c*x^2 - 35*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^4 +
a*x^2), 1/70*(315*(a^3*b^2*c*x^4 + a^4*b*c*x^2)*sqrt(-a*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(-a*c)/(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)) + (10*b^4*c*x^8 + 58*a*b^3*c*x^6 + 156*a^2*b^2*c*x^
4 + 388*a^3*b*c*x^2 - 35*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^4 + a*x^2)]

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25614, size = 204, normalized size = 1.01 \begin{align*} \frac{1}{70} \,{\left (\frac{315 \, a^{4} c^{2} \arctan \left (\frac{\sqrt{b c x^{2} + a c}}{\sqrt{-a c}}\right )}{\sqrt{-a c}} - \frac{35 \, \sqrt{b c x^{2} + a c} a^{4} c}{b x^{2}} + \frac{2 \,{\left (140 \, \sqrt{b c x^{2} + a c} a^{3} c^{15} + 35 \,{\left (b c x^{2} + a c\right )}^{\frac{3}{2}} a^{2} c^{14} + 14 \,{\left (b c x^{2} + a c\right )}^{\frac{5}{2}} a c^{13} + 5 \,{\left (b c x^{2} + a c\right )}^{\frac{7}{2}} c^{12}\right )}}{c^{14}}\right )} b \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(315*a^4*c^2*arctan(sqrt(b*c*x^2 + a*c)/sqrt(-a*c))/sqrt(-a*c) - 35*sqrt(b*c*x^2 + a*c)*a^4*c/(b*x^2) + 2
*(140*sqrt(b*c*x^2 + a*c)*a^3*c^15 + 35*(b*c*x^2 + a*c)^(3/2)*a^2*c^14 + 14*(b*c*x^2 + a*c)^(5/2)*a*c^13 + 5*(
b*c*x^2 + a*c)^(7/2)*c^12)/c^14)*b*sgn(b*x^2 + a)