3.4.29 \(\int \frac {1}{(a+b x^2)^{9/4} (c+d x^2)} \, dx\) [329]

Optimal. Leaf size=274 \[ \frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}+\frac {2 \sqrt {b} (3 b c-8 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} (b c-a d)^2 \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(-b c+a d)^{5/2} x}-\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(-b c+a d)^{5/2} x} \]

[Out]

2/5*b*x/a/(-a*d+b*c)/(b*x^2+a)^(5/4)+2/5*(-8*a*d+3*b*c)*(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^(3
/2)/(-a*d+b*c)^2/(b*x^2+a)^(1/4)+a^(1/4)*d^(3/2)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)
^(1/2),I)*(-b*x^2/a)^(1/2)/(a*d-b*c)^(5/2)/x-a^(1/4)*d^(3/2)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2
)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/(a*d-b*c)^(5/2)/x

________________________________________________________________________________________

Rubi [A]
time = 0.26, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {425, 541, 544, 235, 233, 202, 408, 504, 1232} \begin {gather*} \frac {2 \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} (3 b c-8 a d) E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt [4]{a+b x^2} (b c-a d)^2}+\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{5/2}}-\frac {\sqrt [4]{a} d^{3/2} \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (a d-b c)^{5/2}}+\frac {2 b x}{5 a \left (a+b x^2\right )^{5/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(9/4)*(c + d*x^2)),x]

[Out]

(2*b*x)/(5*a*(b*c - a*d)*(a + b*x^2)^(5/4)) + (2*Sqrt[b]*(3*b*c - 8*a*d)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTa
n[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*(b*c - a*d)^2*(a + b*x^2)^(1/4)) + (a^(1/4)*d^(3/2)*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)^
(5/2)*x) - (a^(1/4)*d^(3/2)*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)^(5/2)*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 233

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

Rule 235

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + b*(x^2
/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

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 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

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*} \int \frac {1}{\left (a+b x^2\right )^{9/4} \left (c+d x^2\right )} \, dx &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}-\frac {2 \int \frac {\frac {1}{2} (-3 b c+5 a d)-\frac {3}{2} b d x^2}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )} \, dx}{5 a (b c-a d)}\\ &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}+\frac {2 b (3 b c-8 a d) x}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}}+\frac {4 \int \frac {\frac {1}{4} \left (-3 b^2 c^2+8 a b c d+5 a^2 d^2\right )-\frac {1}{4} b d (3 b c-8 a d) x^2}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{5 a^2 (b c-a d)^2}\\ &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}+\frac {2 b (3 b c-8 a d) x}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}}+\frac {d^2 \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{(b c-a d)^2}-\frac {(b (3 b c-8 a d)) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{5 a^2 (b c-a d)^2}\\ &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}+\frac {2 b (3 b c-8 a d) x}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}}+\frac {\left (2 d^2 \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)^2 x}-\frac {\left (b (3 b c-8 a d) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}-\frac {\left (d^{3/2} \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)^2 x}+\frac {\left (d^{3/2} \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)^2 x}+\frac {\left (b (3 b c-8 a d) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 b x}{5 a (b c-a d) \left (a+b x^2\right )^{5/4}}+\frac {2 \sqrt {b} (3 b c-8 a d) \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 )}{5 a^{3/2} (b c-a d)^2 \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} d^{3/2} \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)^{5/2} x}-\frac {\sqrt [4]{a} d^{3/2} \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)^{5/2} x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.50, size = 419, normalized size = 1.53 \begin {gather*} \frac {x \left (\frac {b d (-3 b c+8 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} 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 (3 a c \left (5 a^3 d^2+3 b^3 c x^2 \left (c+2 d x^2\right )-a^2 b d \left (10 c+13 d x^2\right )+a b^2 \left (5 c^2-16 d^2 x^4\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 )+b x^2 \left (c+d x^2\right ) \left (9 a^2 d-3 b^2 c x^2-4 a b \left (c-2 d x^2\right )\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 )\right )}{\left (a+b x^2\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 )}{15 a^2 (b c-a d)^2 \sqrt [4]{a+b x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(9/4)*(c + d*x^2)),x]

[Out]

(x*((b*d*(-3*b*c + 8*a*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*(5*a^3*d^2 + 3*b^3*c*x^2*(c + 2*d*x^2) - a^2*b*d*(10*c + 13*d*x^2) + a*b^2*(5*c^2 - 16*d^2*x^4))*A
ppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + b*x^2*(c + d*x^2)*(9*a^2*d - 3*b^2*c*x^2 - 4*a*b*(c -
2*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)])))/((a + b*x^2)*(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)])))))/(15*a^2*(b*c - a*d)^2*(a + b*x^2)^(1/4))

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {9}{4}} \left (d \,x^{2}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(9/4)/(d*x^2+c),x)

[Out]

int(1/(b*x^2+a)^(9/4)/(d*x^2+c),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(9/4)*(d*x^2 + c)), x)

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{2}\right )^{\frac {9}{4}} \left (c + d x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(9/4)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(9/4)*(c + d*x**2)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(9/4)*(d*x^2 + c)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^2+a\right )}^{9/4}\,\left (d\,x^2+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^2)^(9/4)*(c + d*x^2)),x)

[Out]

int(1/((a + b*x^2)^(9/4)*(c + d*x^2)), x)

________________________________________________________________________________________