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

Optimal. Leaf size=308 \[ \frac{3 a c^{5/2} \log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}+\frac{3 a c^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}+1\right )}{4 \sqrt{2} b^{7/4}}-\frac{c (c x)^{3/2} \sqrt [4]{a-b x^2}}{2 b} \]

[Out]

-(c*(c*x)^(3/2)*(a - b*x^2)^(1/4))/(2*b) - (3*a*c^(5/2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b
*x^2)^(1/4))])/(4*Sqrt[2]*b^(7/4)) + (3*a*c^(5/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^
(1/4))])/(4*Sqrt[2]*b^(7/4)) + (3*a*c^(5/2)*Log[Sqrt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/
4)*Sqrt[c*x])/(a - b*x^2)^(1/4)])/(8*Sqrt[2]*b^(7/4)) - (3*a*c^(5/2)*Log[Sqrt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a
- b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)])/(8*Sqrt[2]*b^(7/4))

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Rubi [A]  time = 0.261505, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {321, 329, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 a c^{5/2} \log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}+\frac{3 a c^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}+1\right )}{4 \sqrt{2} b^{7/4}}-\frac{c (c x)^{3/2} \sqrt [4]{a-b x^2}}{2 b} \]

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[1 - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b
*x^2)^(1/4))])/(4*Sqrt[2]*b^(7/4)) + (3*a*c^(5/2)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^
(1/4))])/(4*Sqrt[2]*b^(7/4)) + (3*a*c^(5/2)*Log[Sqrt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/
4)*Sqrt[c*x])/(a - b*x^2)^(1/4)])/(8*Sqrt[2]*b^(7/4)) - (3*a*c^(5/2)*Log[Sqrt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a
- b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)])/(8*Sqrt[2]*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 297

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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{(3 a c) \operatorname{Subst}\left (\int \frac{c-\sqrt{b} x^2}{1+\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{4 b^{3/2}}+\frac{(3 a c) \operatorname{Subst}\left (\int \frac{c+\sqrt{b} x^2}{1+\frac{b x^4}{c^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{\left (3 a c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{c}}{\sqrt [4]{b}}+2 x}{-\frac{c}{\sqrt{b}}-\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}+\frac{\left (3 a c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{c}}{\sqrt [4]{b}}-2 x}{-\frac{c}{\sqrt{b}}+\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}+\frac{\left (3 a c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{c}{\sqrt{b}}-\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 b^2}+\frac{\left (3 a c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{c}{\sqrt{b}}+\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 b^2}\\ &=-\frac{c (c x)^{3/2} \sqrt [4]{a-b x^2}}{2 b}+\frac{3 a c^{5/2} \log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}+\frac{\left (3 a c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}-\frac{\left (3 a c^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}\\ &=-\frac{c (c x)^{3/2} \sqrt [4]{a-b x^2}}{2 b}-\frac{3 a c^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}+\frac{3 a c^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{4 \sqrt{2} b^{7/4}}+\frac{3 a c^{5/2} \log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}-\frac{3 a c^{5/2} \log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{8 \sqrt{2} b^{7/4}}\\ \end{align*}

Mathematica [A]  time = 0.0562795, size = 112, normalized size = 0.36 \[ -\frac{(c x)^{5/2} \left (2 b x^{3/2} \sqrt [4]{a-b x^2}-3 a \sqrt [4]{-b} \tan ^{-1}\left (\frac{\sqrt [4]{-b} \sqrt{x}}{\sqrt [4]{a-b x^2}}\right )+3 a \sqrt [4]{-b} \tanh ^{-1}\left (\frac{\sqrt [4]{-b} \sqrt{x}}{\sqrt [4]{a-b x^2}}\right )\right )}{4 b^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.02, 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 = 36.5276, size = 46, normalized size = 0.15 \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^{2 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(2*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)