3.4.28 \(\int \frac {1}{(a+b x^2)^{7/4} (c+d x^2)} \, dx\) [328]

Optimal. Leaf size=254 \[ \frac {2 b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/4}}+\frac {2 \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} (b c-a d) \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} d \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(b c-a d)^2 x}-\frac {\sqrt [4]{a} d \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{(b c-a d)^2 x} \]

[Out]

2/3*b*x/a/(-a*d+b*c)/(b*x^2+a)^(3/4)+2/3*(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)))*EllipticF(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
)^(3/4)/a^(1/2)-a^(1/4)*d*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)^2/x-a^(1/4)*d*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)^2/x

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Rubi [A]
time = 0.11, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {412, 205, 239, 237, 410, 109, 418, 1232} \begin {gather*} -\frac {\sqrt [4]{a} d \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 (b c-a d)^2}-\frac {\sqrt [4]{a} d \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 (b c-a d)^2}+\frac {2 \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \left (a+b x^2\right )^{3/4} (b c-a d)}+\frac {2 b x}{3 a \left (a+b x^2\right )^{3/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)^(7/4)*(c + d*x^2)),x]

[Out]

(2*b*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/4)) + (2*Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sq
rt[a]]/2, 2])/(3*Sqrt[a]*(b*c - a*d)*(a + b*x^2)^(3/4)) - (a^(1/4)*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)^2*x) - (a^(1/4)*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
)^2*x)

Rule 109

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 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 237

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

Rule 239

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 410

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 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 418

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 9.77, size = 331, normalized size = 1.30 \begin {gather*} \frac {x \left (-\frac {b d x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} 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}+\frac {6 \left (3 a c \left (-3 b c+3 a d-2 b d x^2\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 )+b x^2 \left (c+d 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 )\right )}{\left (c+d x^2\right ) \left (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 )-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 )\right )}\right )}{9 a (-b c+a d) \left (a+b x^2\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)^(7/4)*(c + d*x^2)),x]

[Out]

(x*(-((b*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*(-
3*b*c + 3*a*d - 2*b*d*x^2)*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + b*x^2*(c + d*x^2)*(4*a*d*A
ppellF1[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)])))/((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)])))))
/(9*a*(-(b*c) + a*d)*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {7}{4}} \left (d \,x^{2}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(7/4)/(d*x^2+c),x)

[Out]

int(1/(b*x^2+a)^(7/4)/(d*x^2+c),x)

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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)^(7/4)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(7/4)*(d*x^2 + c)), x)

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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)^(7/4)/(d*x^2+c),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x^{2}\right )^{\frac {7}{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)**(7/4)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(7/4)*(c + d*x**2)), x)

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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)^(7/4)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(7/4)*(d*x^2 + c)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (b\,x^2+a\right )}^{7/4}\,\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)^(7/4)*(c + d*x^2)),x)

[Out]

int(1/((a + b*x^2)^(7/4)*(c + d*x^2)), x)

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