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

Optimal. Leaf size=117 \[ \frac {2}{3} \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \tan ^{-1}\left (\sqrt [4]{\frac {b x^2}{a}+1}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}}-\frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \tanh ^{-1}\left (\sqrt [4]{\frac {b x^2}{a}+1}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4}} \]

[Out]

2/3*(c*(b*x^2+a)^(1/2))^(3/2)+arctan((1+b*x^2/a)^(1/4))*(c*(b*x^2+a)^(1/2))^(3/2)/(1+b*x^2/a)^(3/4)-arctanh((1
+b*x^2/a)^(1/4))*(c*(b*x^2+a)^(1/2))^(3/2)/(1+b*x^2/a)^(3/4)

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Rubi [A]  time = 0.16, antiderivative size = 141, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6720, 266, 50, 63, 298, 203, 206} \[ \frac {a^{3/4} c \sqrt {c \sqrt {a+b x^2}} \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a+b x^2}}-\frac {a^{3/4} c \sqrt {c \sqrt {a+b x^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a+b x^2}}+\frac {2}{3} c \sqrt {a+b x^2} \sqrt {c \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2)/x,x]

[Out]

(2*c*Sqrt[c*Sqrt[a + b*x^2]]*Sqrt[a + b*x^2])/3 + (a^(3/4)*c*Sqrt[c*Sqrt[a + b*x^2]]*ArcTan[(a + b*x^2)^(1/4)/
a^(1/4)])/(a + b*x^2)^(1/4) - (a^(3/4)*c*Sqrt[c*Sqrt[a + b*x^2]]*ArcTanh[(a + b*x^2)^(1/4)/a^(1/4)])/(a + b*x^
2)^(1/4)

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 203

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

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

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 \sqrt {a+b x^2}\right )^{3/2}}{x} \, dx &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {\left (a+b x^2\right )^{3/4}}{x} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/4}}{x} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac {2}{3} c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {\left (a c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt [4]{a+b x}} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac {2}{3} c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {\left (2 a c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{b \sqrt [4]{a+b x^2}}\\ &=\frac {2}{3} c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}-\frac {\left (a c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^2}\right )}{\sqrt [4]{a+b x^2}}+\frac {\left (a c \sqrt {c \sqrt {a+b x^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^2}\right )}{\sqrt [4]{a+b x^2}}\\ &=\frac {2}{3} c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}+\frac {a^{3/4} c \sqrt {c \sqrt {a+b x^2}} \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a+b x^2}}-\frac {a^{3/4} c \sqrt {c \sqrt {a+b x^2}} \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 96, normalized size = 0.82 \[ \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \left (3 a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )-3 a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )+2 \left (a+b x^2\right )^{3/4}\right )}{3 \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*Sqrt[a + b*x^2])^(3/2)*(2*(a + b*x^2)^(3/4) + 3*a^(3/4)*ArcTan[(a + b*x^2)^(1/4)/a^(1/4)] - 3*a^(3/4)*ArcT
anh[(a + b*x^2)^(1/4)/a^(1/4)]))/(3*(a + b*x^2)^(3/4))

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fricas [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)^(1/2))^(3/2)/x,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.41, size = 190, normalized size = 1.62 \[ -\frac {1}{12} \, {\left (6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + 6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) + 3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{2} + a} + \sqrt {-a}\right ) - 8 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}}\right )} c^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x,x, algorithm="giac")

[Out]

-1/12*(6*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^2 + a)^(1/4))/(-a)^(1/4)) + 6*sqrt
(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^2 + a)^(1/4))/(-a)^(1/4)) - 3*sqrt(2)*(-a)^(3/
4)*log(sqrt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a)) + 3*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*
(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a)) - 8*(b*x^2 + a)^(3/4))*c^(3/2)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sqrt {b \,x^{2}+a}\, c \right )^{\frac {3}{2}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)^(1/2)*c)^(3/2)/x,x)

[Out]

int(((b*x^2+a)^(1/2)*c)^(3/2)/x,x)

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maxima [A]  time = 1.97, size = 118, normalized size = 1.01 \[ \frac {3 \, a c^{4} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\sqrt {b x^{2} + a} c}}{\left (a c^{2}\right )^{\frac {1}{4}}}\right )}{\left (a c^{2}\right )^{\frac {1}{4}}} + \frac {\log \left (\frac {\sqrt {\sqrt {b x^{2} + a} c} - \left (a c^{2}\right )^{\frac {1}{4}}}{\sqrt {\sqrt {b x^{2} + a} c} + \left (a c^{2}\right )^{\frac {1}{4}}}\right )}{\left (a c^{2}\right )^{\frac {1}{4}}}\right )} + 4 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} c^{2}}{6 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x^2)^(1/2))^(3/2)/x,x)

[Out]

int((c*(a + b*x^2)^(1/2))^(3/2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x**2+a)**(1/2))**(3/2)/x,x)

[Out]

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

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