3.327 \(\int \frac{1}{(a+b x^2)^{5/4} (c+d x^2)} \, dx\)

Optimal. Leaf size=233 \[ \frac{2 \sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt [4]{a+b x^2} (b c-a d)}+\frac{\sqrt [4]{a} \sqrt{d} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{3/2}}-\frac{\sqrt [4]{a} \sqrt{d} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{3/2}} \]

[Out]

(2*Sqrt[b]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*(b*c - a*d)*(a + b*x^2)
^(1/4)) + (a^(1/4)*Sqrt[d]*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a +
b*x^2)^(1/4)/a^(1/4)], -1])/((-(b*c) + a*d)^(3/2)*x) - (a^(1/4)*Sqrt[d]*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]
*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/((-(b*c) + a*d)^(3/2)*x)

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Rubi [A]  time = 0.160015, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {403, 197, 196, 399, 490, 1218} \[ \frac{2 \sqrt{b} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \sqrt [4]{a+b x^2} (b c-a d)}+\frac{\sqrt [4]{a} \sqrt{d} \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{3/2}}-\frac{\sqrt [4]{a} \sqrt{d} \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(5/4)*(c + d*x^2)),x]

[Out]

(2*Sqrt[b]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*(b*c - a*d)*(a + b*x^2)
^(1/4)) + (a^(1/4)*Sqrt[d]*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a +
b*x^2)^(1/4)/a^(1/4)], -1])/((-(b*c) + a*d)^(3/2)*x) - (a^(1/4)*Sqrt[d]*Sqrt[-((b*x^2)/a)]*EllipticPi[(Sqrt[a]
*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/((-(b*c) + a*d)^(3/2)*x)

Rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/(b*c - a*d), Int[(a + b*x^2)^p, x],
x] - Dist[d/(b*c - a*d), Int[(a + b*x^2)^(p + 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*
d, 0] && LtQ[p, -1] && EqQ[Denominator[p], 4] && (EqQ[p, -5/4] || EqQ[p, -7/4])

Rule 197

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + (b
*x^2)/a)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

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

Mathematica [C]  time = 0.268941, size = 327, normalized size = 1.4 \[ \frac{x \left (\frac{b d x^2 \sqrt [4]{\frac{b x^2}{a}+1} F_1\left (\frac{3}{2};\frac{1}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{c}+\frac{6 \left (b x^2 \left (c+d x^2\right ) \left (4 a d F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+b c F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )+3 a c \left (a d-b \left (c+2 d x^2\right )\right ) F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )}{\left (c+d x^2\right ) \left (6 a c F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-x^2 \left (4 a d F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+b c F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )\right )}\right )}{3 a \sqrt [4]{a+b x^2} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(5/4)*(c + d*x^2)),x]

[Out]

(x*((b*d*x^2*(1 + (b*x^2)/a)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/c + (6*(3*a*c*(a*d
- b*(c + 2*d*x^2))*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + b*x^2*(c + d*x^2)*(4*a*d*AppellF1[
3/2, 1/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/
((c + d*x^2)*(6*a*c*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] - x^2*(4*a*d*AppellF1[3/2, 1/4, 2,
5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))))/(3*a*(-(b*c
) + a*d)*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(5/4)/(d*x^2+c),x)

[Out]

int(1/(b*x^2+a)^(5/4)/(d*x^2+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}{\left (d x^{2} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)), 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(1/(b*x^2+a)^(5/4)/(d*x^2+c),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right )^{\frac{5}{4}} \left (c + d x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(5/4)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(5/4)*(c + d*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}{\left (d x^{2} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)), x)