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]

3/2*a^2*b*c*(c*(b*x^2+a)^3)^(1/2)+9/2*a^3*b*c*(c*(b*x^2+a)^3)^(1/2)/(b*x^2+a)+9/10*a*b*c*(b*x^2+a)*(c*(b*x^2+a
)^3)^(1/2)+9/14*b*c*(b*x^2+a)^2*(c*(b*x^2+a)^3)^(1/2)-1/2*c*(b*x^2+a)^3*(c*(b*x^2+a)^3)^(1/2)/x^2-9/2*a^2*b*c*
arctanh((1+b*x^2/a)^(1/2))*(c*(b*x^2+a)^3)^(1/2)/(1+b*x^2/a)^(3/2)

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Rubi [A]  time = 0.22, 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 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]

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 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])

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*}

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Mathematica [C]  time = 0.02, 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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fricas [A]  time = 0.47, size = 411, normalized size = 2.03 \[ \left [\frac {315 \, {\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt {a c} \log \left (-\frac {b^{2} c x^{4} + 3 \, a b c x^{2} + 2 \, a^{2} c - 2 \, \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {a c}}{b x^{4} + a x^{2}}\right ) + 2 \, {\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{140 \, {\left (b x^{4} + a x^{2}\right )}}, \frac {315 \, {\left (a^{3} b^{2} c x^{4} + a^{4} b c x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c} \sqrt {-a c}}{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}\right ) + {\left (10 \, b^{4} c x^{8} + 58 \, a b^{3} c x^{6} + 156 \, a^{2} b^{2} c x^{4} + 388 \, a^{3} b c x^{2} - 35 \, a^{4} c\right )} \sqrt {b^{3} c x^{6} + 3 \, a b^{2} c x^{4} + 3 \, a^{2} b c x^{2} + a^{3} c}}{70 \, {\left (b x^{4} + a x^{2}\right )}}\right ] \]

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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giac [A]  time = 0.39, size = 204, normalized size = 1.01 \[ \frac {{\left (\frac {315 \, a^{4} b^{2} c \arctan \left (\frac {\sqrt {b c x^{2} + a c}}{\sqrt {-a c}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {-a c}} - \frac {35 \, \sqrt {b c x^{2} + a c} a^{4} b \mathrm {sgn}\left (b x^{2} + a\right )}{x^{2}} + \frac {2 \, {\left (140 \, \sqrt {b c x^{2} + a c} a^{3} b^{2} c^{21} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, {\left (b c x^{2} + a c\right )}^{\frac {3}{2}} a^{2} b^{2} c^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + 14 \, {\left (b c x^{2} + a c\right )}^{\frac {5}{2}} a b^{2} c^{19} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, {\left (b c x^{2} + a c\right )}^{\frac {7}{2}} b^{2} c^{18} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{c^{21}}\right )} c}{70 \, b} \]

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*b^2*c*arctan(sqrt(b*c*x^2 + a*c)/sqrt(-a*c))*sgn(b*x^2 + a)/sqrt(-a*c) - 35*sqrt(b*c*x^2 + a*c)*
a^4*b*sgn(b*x^2 + a)/x^2 + 2*(140*sqrt(b*c*x^2 + a*c)*a^3*b^2*c^21*sgn(b*x^2 + a) + 35*(b*c*x^2 + a*c)^(3/2)*a
^2*b^2*c^20*sgn(b*x^2 + a) + 14*(b*c*x^2 + a*c)^(5/2)*a*b^2*c^19*sgn(b*x^2 + a) + 5*(b*c*x^2 + a*c)^(7/2)*b^2*
c^18*sgn(b*x^2 + a))/c^21)*c/b

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maple [A]  time = 0.02, size = 238, normalized size = 1.18 \[ \frac {\left (\left (b \,x^{2}+a \right )^{3} c \right )^{\frac {3}{2}} \left (-315 a^{4} b \,c^{3} x^{2} \ln \left (\frac {2 a c +2 \sqrt {a c}\, \sqrt {b c \,x^{2}+a c}}{x}\right )+315 \sqrt {b c \,x^{2}+a c}\, \sqrt {a c}\, a^{3} b \,c^{2} x^{2}+105 \left (b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, a^{2} b c \,x^{2}+10 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} b^{2} x^{4}-4 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a b \,x^{2}+42 \left (\left (b \,x^{2}+a \right ) c \right )^{\frac {5}{2}} \sqrt {a c}\, a b \,x^{2}-35 \sqrt {a c}\, \left (b c \,x^{2}+a c \right )^{\frac {5}{2}} a^{2}\right )}{70 \left (b \,x^{2}+a \right )^{3} \left (\left (b \,x^{2}+a \right ) c \right )^{\frac {3}{2}} \sqrt {a c}\, c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left ({\left (b x^{2} + a\right )}^{3} c\right )^{\frac {3}{2}}}{x^{3}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,{\left (b\,x^2+a\right )}^3\right )}^{3/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x^2)^3)^(3/2)/x^3,x)

[Out]

int((c*(a + b*x^2)^3)^(3/2)/x^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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