3.330 \(\int \frac {1}{(a+b x^2)^{11/4} (c+d x^2)} \, dx\)

Optimal. Leaf size=304 \[ \frac {2 \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 a^{3/2} \left (a+b x^2\right )^{3/4} (b c-a d)^2}+\frac {2 b x (5 b c-12 a d)}{21 a^2 \left (a+b x^2\right )^{3/4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)^3}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)} \]

[Out]

2/7*b*x/a/(-a*d+b*c)/(b*x^2+a)^(7/4)+2/21*b*(-12*a*d+5*b*c)*x/a^2/(-a*d+b*c)^2/(b*x^2+a)^(3/4)+2/21*(-12*a*d+5
*b*c)*(1+b*x^2/a)^(3/4)*(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)))*Ellipt
icF(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)^(3/4)+a^(1/4)*d^2*Ellip
ticPi((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)^3/x+a^(1/4)*d^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)^3/x

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Rubi [A]  time = 0.34, antiderivative size = 304, 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 = {414, 527, 530, 233, 231, 401, 108, 409, 1218} \[ \frac {2 b x (5 b c-12 a d)}{21 a^2 \left (a+b x^2\right )^{3/4} (b c-a d)^2}+\frac {2 \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (5 b c-12 a d) F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} \left (a+b x^2\right )^{3/4} (b c-a d)^2}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)^3}+\frac {\sqrt [4]{a} d^2 \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{x (b c-a d)^3}+\frac {2 b x}{7 a \left (a+b x^2\right )^{7/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(11/4)*(c + d*x^2)),x]

[Out]

(2*b*x)/(7*a*(b*c - a*d)*(a + b*x^2)^(7/4)) + (2*b*(5*b*c - 12*a*d)*x)/(21*a^2*(b*c - a*d)^2*(a + b*x^2)^(3/4)
) + (2*Sqrt[b]*(5*b*c - 12*a*d)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*a^(3/2)
*(b*c - a*d)^2*(a + b*x^2)^(3/4)) + (a^(1/4)*d^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)^3*x) + (a^(1/4)*d^2*Sqrt[-((b*x^2)/a)]*Elliptic
Pi[(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*x)

Rule 108

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - (d*e)/f + (d*x^4)/f]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-(f/(d*e - c*f)), 0]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 401

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[Sqrt[-((b*x^2)/a)]/(2*x), Subst[I
nt[1/(Sqrt[-((b*x)/a)]*(a + b*x)^(3/4)*(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0]

Rule 409

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

Rule 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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 530

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 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 )^{11/4} \left (c+d x^2\right )} \, dx &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}-\frac {2 \int \frac {\frac {1}{2} (-5 b c+7 a d)-\frac {5}{2} b d x^2}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )} \, dx}{7 a (b c-a d)}\\ &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {4 \int \frac {\frac {1}{4} \left (5 b^2 c^2-12 a b c d+21 a^2 d^2\right )+\frac {1}{4} b d (5 b c-12 a d) x^2}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx}{21 a^2 (b c-a d)^2}\\ &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {d^2 \int \frac {1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx}{(b c-a d)^2}+\frac {(b (5 b c-12 a d)) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{21 a^2 (b c-a d)^2}\\ &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {\left (d^2 \sqrt {-\frac {b x^2}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-\frac {b x}{a}} (a+b x)^{3/4} (c+d x)} \, dx,x,x^2\right )}{2 (b c-a d)^2 x}+\frac {\left (b (5 b c-12 a d) \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {b} (5 b c-12 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac {\left (2 d^2 \sqrt {-\frac {b x^2}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\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 {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {b} (5 b c-12 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {\left (d^2 \sqrt {-\frac {b x^2}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {-b c+a d}}\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{(b c-a d)^3 x}+\frac {\left (d^2 \sqrt {-\frac {b x^2}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {-b c+a d}}\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{(b c-a d)^3 x}\\ &=\frac {2 b x}{7 a (b c-a d) \left (a+b x^2\right )^{7/4}}+\frac {2 b (5 b c-12 a d) x}{21 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {b} (5 b c-12 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} d^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)^3 x}+\frac {\sqrt [4]{a} d^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)^3 x}\\ \end {align*}

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Mathematica [C]  time = 0.78, size = 431, normalized size = 1.42 \[ -\frac {x \left (\frac {6 \left (b x^2 \left (c+d x^2\right ) \left (15 a^2 d+a b \left (12 d x^2-8 c\right )-5 b^2 c x^2\right ) \left (4 a d F_1\left (\frac {3}{2};\frac {3}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c F_1\left (\frac {3}{2};\frac {7}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )+3 a c \left (21 a^3 d^2-3 a^2 b d \left (14 c+3 d x^2\right )+a b^2 \left (21 c^2-20 c d x^2-24 d^2 x^4\right )+5 b^3 c x^2 \left (3 c+2 d x^2\right )\right ) F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{\left (a+b x^2\right ) \left (c+d x^2\right ) \left (x^2 \left (4 a d F_1\left (\frac {3}{2};\frac {3}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c F_1\left (\frac {3}{2};\frac {7}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )-6 a c F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}+\frac {b d x^2 \left (\frac {b x^2}{a}+1\right )^{3/4} (12 a d-5 b c) F_1\left (\frac {3}{2};\frac {3}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}\right )}{63 a^2 \left (a+b x^2\right )^{3/4} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(11/4)*(c + d*x^2)),x]

[Out]

-1/63*(x*((b*d*(-5*b*c + 12*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/
c)])/c + (6*(3*a*c*(21*a^3*d^2 + 5*b^3*c*x^2*(3*c + 2*d*x^2) - 3*a^2*b*d*(14*c + 3*d*x^2) + a*b^2*(21*c^2 - 20
*c*d*x^2 - 24*d^2*x^4))*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + b*x^2*(c + d*x^2)*(15*a^2*d -
5*b^2*c*x^2 + a*b*(-8*c + 12*d*x^2))*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*Ap
pellF1[3/2, 7/4, 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, 3/4, 1,
3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*
AppellF1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))))/(a^2*(b*c - a*d)^2*(a + b*x^2)^(3/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4)/(d*x^2+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{4}} {\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(11/4)*(d*x^2 + c)), x)

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {11}{4}} \left (d \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(11/4)/(d*x^2+c),x)

[Out]

int(1/(b*x^2+a)^(11/4)/(d*x^2+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{4}} {\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(11/4)*(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^2+a\right )}^{11/4}\,\left (d\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^2)^(11/4)*(c + d*x^2)),x)

[Out]

int(1/((a + b*x^2)^(11/4)*(c + d*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{2}\right )^{\frac {11}{4}} \left (c + d x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(11/4)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(11/4)*(c + d*x**2)), x)

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