3.944 \(\int \frac{(c x)^{3/2}}{\sqrt [4]{a+b x^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0648402, 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, 240, 212, 208, 205} \[ -\frac{a c^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}}-\frac{a c^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}}+\frac{c \sqrt{c x} \left (a+b x^2\right )^{3/4}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/(a + b*x^2)^(1/4),x]

[Out]

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

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 240

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

Rule 212

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0362232, size = 97, normalized size = 0.83 \[ \frac{(c x)^{3/2} \left (2 \sqrt [4]{b} \sqrt{x} \left (a+b x^2\right )^{3/4}-a \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )-a \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{5/4} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(3/2)/(a + b*x^2)^(1/4),x]

[Out]

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

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)/(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(3/2)/(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(3/2)/(b*x^2 + a)^(1/4), x)

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Fricas [B]  time = 1.79081, size = 689, normalized size = 5.89 \begin{align*} \frac{4 \,{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} c + 4 \, \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{1}{4}} b \arctan \left (-\frac{\left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{3}{4}}{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} a b^{4} c -{\left (b^{5} x^{2} + a b^{4}\right )} \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{3}{4}} \sqrt{\frac{\sqrt{b x^{2} + a} a^{2} c^{3} x + \sqrt{\frac{a^{4} c^{6}}{b^{5}}}{\left (b^{3} x^{2} + a b^{2}\right )}}{b x^{2} + a}}}{a^{4} b c^{6} x^{2} + a^{5} c^{6}}\right ) - \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{1}{4}} b \log \left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} a c + \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{1}{4}}{\left (b^{2} x^{2} + a b\right )}}{b x^{2} + a}\right ) + \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{1}{4}} b \log \left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} a c - \left (\frac{a^{4} c^{6}}{b^{5}}\right )^{\frac{1}{4}}{\left (b^{2} x^{2} + a b\right )}}{b x^{2} + a}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

1/8*(4*(b*x^2 + a)^(3/4)*sqrt(c*x)*c + 4*(a^4*c^6/b^5)^(1/4)*b*arctan(-((a^4*c^6/b^5)^(3/4)*(b*x^2 + a)^(3/4)*
sqrt(c*x)*a*b^4*c - (b^5*x^2 + a*b^4)*(a^4*c^6/b^5)^(3/4)*sqrt((sqrt(b*x^2 + a)*a^2*c^3*x + sqrt(a^4*c^6/b^5)*
(b^3*x^2 + a*b^2))/(b*x^2 + a)))/(a^4*b*c^6*x^2 + a^5*c^6)) - (a^4*c^6/b^5)^(1/4)*b*log(((b*x^2 + a)^(3/4)*sqr
t(c*x)*a*c + (a^4*c^6/b^5)^(1/4)*(b^2*x^2 + a*b))/(b*x^2 + a)) + (a^4*c^6/b^5)^(1/4)*b*log(((b*x^2 + a)^(3/4)*
sqrt(c*x)*a*c - (a^4*c^6/b^5)^(1/4)*(b^2*x^2 + a*b))/(b*x^2 + a)))/b

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Sympy [C]  time = 3.63552, size = 44, normalized size = 0.38 \begin{align*} \frac{c^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(3/2)/(b*x**2+a)**(1/4),x)

[Out]

c**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(1/4)*gamma(9/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((c*x)^(3/2)/(b*x^2 + a)^(1/4), x)