\(\int \frac {1}{(a+b x^2)^{5/4} (c+d x^2)} \, dx\) [327]

Optimal result

Integrand size = 21, antiderivative size = 233 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\frac {2 \sqrt {b} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} (b c-a d) \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} \sqrt {d} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(-b c+a d)^{3/2} x}-\frac {\sqrt [4]{a} \sqrt {d} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}},\arcsin \left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right ),-1\right )}{(-b c+a d)^{3/2} x} \]

[Out]

2*(1+b*x^2/a)^(1/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE(
sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/(-a*d+b*c)/(b*x^2+a)^(1/4)/a^(1/2)+a^(1/4)*EllipticPi((b*x
^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*d^(1/2)*(-b*x^2/a)^(1/2)/(a*d-b*c)^(3/2)/x-a^(1/4)*Ell
ipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*d^(1/2)*(-b*x^2/a)^(1/2)/(a*d-b*c)^(3/2)/x

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {412, 203, 202, 408, 504, 1232} \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\frac {\sqrt [4]{a} \sqrt {d} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{x (a d-b c)^{3/2}}-\frac {\sqrt [4]{a} \sqrt {d} \sqrt {-\frac {b x^2}{a}} \operatorname {EllipticPi}\left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}},\arcsin \left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right ),-1\right )}{x (a d-b c)^{3/2}}+\frac {2 \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a+b x^2} (b c-a d)} \]

[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 202

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

Rule 203

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 408

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 412

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 504

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), Int[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 1232

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

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 10.49 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\frac {x \left (\frac {b d x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}+\frac {6 \left (3 a c \left (a d-b \left (c+2 d x^2\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b x^2 \left (c+d x^2\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (4 a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{3 a (-b c+a d) \sqrt [4]{a+b x^2}} \]

[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))

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (d \,x^{2}+c \right )}d x\]

[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)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/4}\,\left (d\,x^2+c\right )} \,d x \]

[In]

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

[Out]

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