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

Optimal. Leaf size=133 \[ -\frac{\left (c \sqrt{a+b x^2}\right )^{3/2}}{2 x^2}+\frac{3 b \left (c \sqrt{a+b x^2}\right )^{3/2} \tan ^{-1}\left (\sqrt [4]{\frac{b x^2}{a}+1}\right )}{4 a \left (\frac{b x^2}{a}+1\right )^{3/4}}-\frac{3 b \left (c \sqrt{a+b x^2}\right )^{3/2} \tanh ^{-1}\left (\sqrt [4]{\frac{b x^2}{a}+1}\right )}{4 a \left (\frac{b x^2}{a}+1\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.162866, antiderivative size = 151, normalized size of antiderivative = 1.14, 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, 47, 63, 298, 203, 206} \[ -\frac{c \sqrt{a+b x^2} \sqrt{c \sqrt{a+b x^2}}}{2 x^2}+\frac{3 b c \sqrt{c \sqrt{a+b x^2}} \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 \sqrt [4]{a} \sqrt [4]{a+b x^2}}-\frac{3 b c \sqrt{c \sqrt{a+b x^2}} \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 \sqrt [4]{a} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2)/x^3,x]

[Out]

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

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{\left (c \sqrt{a+b x^2}\right )^{3/2}}{x^3} \, dx &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \int \frac{\left (a+b x^2\right )^{3/4}}{x^3} \, 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^2} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{2 x^2}+\frac{\left (3 b 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 )}{8 \sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{2 x^2}+\frac{\left (3 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 )}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{2 x^2}-\frac{\left (3 b 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 )}{4 \sqrt [4]{a+b x^2}}+\frac{\left (3 b 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 )}{4 \sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{2 x^2}+\frac{3 b c \sqrt{c \sqrt{a+b x^2}} \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 \sqrt [4]{a} \sqrt [4]{a+b x^2}}-\frac{3 b c \sqrt{c \sqrt{a+b x^2}} \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )}{4 \sqrt [4]{a} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0126223, size = 50, normalized size = 0.38 \[ \frac{2 b \left (a+b x^2\right ) \left (c \sqrt{a+b x^2}\right )^{3/2} \, _2F_1\left (\frac{7}{4},2;\frac{11}{4};\frac{b x^2}{a}+1\right )}{7 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*Hypergeometric2F1[7/4, 2, 11/4, 1 + (b*x^2)/a])/(7*a^2)

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Maple [F]  time = 0.008, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( c\sqrt{b{x}^{2}+a} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \sqrt{a + b x^{2}}\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)**(1/2))**(3/2)/x**3,x)

[Out]

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

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Giac [A]  time = 1.19769, size = 282, normalized size = 2.12 \begin{align*} -\frac{1}{16} \,{\left (\frac{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 )}{a} + \frac{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 )}{a} - \frac{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 )}{a} + \frac{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 )}{a} + \frac{8 \,{\left (b x^{2} + a\right )}^{\frac{3}{4}}}{b x^{2}}\right )} b c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(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))/a + 6*sq
rt(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^2 + a)^(1/4))/(-a)^(1/4))/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))/a + 3*sqrt(2)*(-a)^(3/4)*log(-sq
rt(2)*(b*x^2 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^2 + a) + sqrt(-a))/a + 8*(b*x^2 + a)^(3/4)/(b*x^2))*b*c^(3/2)