3.970 \(\int \frac{(c x)^{5/2}}{(a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=117 \[ \frac{3 a c^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}-\frac{3 a c^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac{c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b} \]

[Out]

(c*(c*x)^(3/2)*(a + b*x^2)^(1/4))/(2*b) + (3*a*c^(5/2)*ArcTan[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))]
)/(4*b^(7/4)) - (3*a*c^(5/2)*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(4*b^(7/4))

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Rubi [A]  time = 0.0668808, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {321, 329, 331, 298, 205, 208} \[ \frac{3 a c^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}-\frac{3 a c^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{7/4}}+\frac{c (c x)^{3/2} \sqrt [4]{a+b x^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)/(a + b*x^2)^(3/4),x]

[Out]

(c*(c*x)^(3/2)*(a + b*x^2)^(1/4))/(2*b) + (3*a*c^(5/2)*ArcTan[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))]
)/(4*b^(7/4)) - (3*a*c^(5/2)*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(4*b^(7/4))

Rule 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (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 205

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

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

Mathematica [A]  time = 0.037342, size = 97, normalized size = 0.83 \[ \frac{(c x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2}+3 a \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )-3 a \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{7/4} x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(5/2)/(a + b*x^2)^(3/4),x]

[Out]

((c*x)^(5/2)*(2*b^(3/4)*x^(3/2)*(a + b*x^2)^(1/4) + 3*a*ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] - 3*a*ArcT
anh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(4*b^(7/4)*x^(5/2))

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{5}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)/(b*x^2+a)^(3/4),x)

[Out]

int((c*x)^(5/2)/(b*x^2+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(5/2)/(b*x^2 + a)^(3/4), x)

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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*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C]  time = 33.5981, size = 44, normalized size = 0.38 \begin{align*} \frac{c^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/2)/(b*x**2+a)**(3/4),x)

[Out]

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((c*x)^(5/2)/(b*x^2 + a)^(3/4), x)