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

Optimal. Leaf size=107 \[ \frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0594885, antiderivative size = 107, 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 = {288, 329, 240, 212, 208, 205} \[ \frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/(a + b*x^2)^(5/4),x]

[Out]

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

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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}}{\left (a+b x^2\right )^{5/4}} \, dx &=-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}}+\frac{c^2 \int \frac{1}{\sqrt{c x} \sqrt [4]{a+b x^2}} \, dx}{b}\\ &=-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}}+\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{b}\\ &=-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}}+\frac{(2 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 )}{b}\\ &=-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}}+\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{c-\sqrt{b} x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{b}+\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{c+\sqrt{b} x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{b}\\ &=-\frac{2 c \sqrt{c x}}{b \sqrt [4]{a+b x^2}}+\frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}+\frac{c^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{5/4}}\\ \end{align*}

Mathematica [C]  time = 0.0119004, size = 59, normalized size = 0.55 \[ \frac{2 x (c x)^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{5}{4},\frac{5}{4};\frac{9}{4};-\frac{b x^2}{a}\right )}{5 a \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(c*x)^(3/2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((b*x^2)/a)])/(5*a*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{{\frac{3}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B]  time = 1.72883, size = 701, normalized size = 6.55 \begin{align*} -\frac{4 \,{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} c + 4 \,{\left (b^{2} x^{2} + a b\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} b^{4} c \left (\frac{c^{6}}{b^{5}}\right )^{\frac{3}{4}} -{\left (b^{5} x^{2} + a b^{4}\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{3}{4}} \sqrt{\frac{\sqrt{b x^{2} + a} c^{3} x +{\left (b^{3} x^{2} + a b^{2}\right )} \sqrt{\frac{c^{6}}{b^{5}}}}{b x^{2} + a}}}{b c^{6} x^{2} + a c^{6}}\right ) -{\left (b^{2} x^{2} + a b\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{1}{4}} \log \left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} c +{\left (b^{2} x^{2} + a b\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{1}{4}}}{b x^{2} + a}\right ) +{\left (b^{2} x^{2} + a b\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{1}{4}} \log \left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} c -{\left (b^{2} x^{2} + a b\right )} \left (\frac{c^{6}}{b^{5}}\right )^{\frac{1}{4}}}{b x^{2} + a}\right )}{2 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 13.3458, size = 44, normalized size = 0.41 \begin{align*} \frac{c^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{4}} \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)**(5/4),x)

[Out]

c**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*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{5}{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)^(5/4),x, algorithm="giac")

[Out]

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