3.211 \(\int \frac {(a+b x^4)^{9/4}}{(c+d x^4)^2} \, dx\)

Optimal. Leaf size=353 \[ -\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (3 b c-a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 c d^2 \left (a+b x^4\right )^{3/4}}-\frac {3 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) (a d+2 b c) \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}-\frac {3 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) (a d+2 b c) \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}+\frac {b x \sqrt [4]{a+b x^4} (3 b c-a d)}{4 c d^2}-\frac {x \left (a+b x^4\right )^{5/4} (b c-a d)}{4 c d \left (c+d x^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {413, 528, 529, 237, 335, 275, 231, 407, 409, 1218} \[ -\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (3 b c-a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 c d^2 \left (a+b x^4\right )^{3/4}}-\frac {3 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) (a d+2 b c) \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}-\frac {3 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) (a d+2 b c) \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}+\frac {b x \sqrt [4]{a+b x^4} (3 b c-a d)}{4 c d^2}-\frac {x \left (a+b x^4\right )^{5/4} (b c-a d)}{4 c d \left (c+d x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*b*c - a*d)*x*(a + b*x^4)^(1/4))/(4*c*d^2) - ((b*c - a*d)*x*(a + b*x^4)^(5/4))/(4*c*d*(c + d*x^4)) - (Sqr
t[a]*b^(3/2)*(3*b*c - a*d)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*c*d^2*(
a + b*x^4)^(3/4)) - (3*(b*c - a*d)*(2*b*c + a*d)*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]*EllipticPi[-(Sqrt[b*c - a
*d]/(Sqrt[b]*Sqrt[c])), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1])/(8*b^(1/4)*c^2*d^2) - (3*(b*c - a*d)*(2*b*
c + a*d)*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]*EllipticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)*x)/
(a + b*x^4)^(1/4)], -1])/(8*b^(1/4)*c^2*d^2)

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 237

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 407

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[Sqrt[a + b*x^4]*Sqrt[a/(a + b*x^4)],
Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c - a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], 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 413

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

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 529

Int[((e_) + (f_.)*(x_)^4)/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[(b*e - a*f)/(
b*c - a*d), Int[1/(a + b*x^4)^(3/4), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(a + b*x^4)^(1/4)/(c + d*x^4),
x], x] /; FreeQ[{a, b, c, d, e, f}, 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 {\left (a+b x^4\right )^{9/4}}{\left (c+d x^4\right )^2} \, dx &=-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}+\frac {\int \frac {\sqrt [4]{a+b x^4} \left (a (b c+3 a d)+2 b (3 b c-a d) x^4\right )}{c+d x^4} \, dx}{4 c d}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}+\frac {\int \frac {-2 a \left (3 b^2 c^2-2 a b c d-3 a^2 d^2\right )-4 b \left (3 b^2 c^2-3 a b c d-a^2 d^2\right ) x^4}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx}{8 c d^2}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}+\frac {(a b (3 b c-a d)) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{4 c d^2}-\frac {(3 (b c-a d) (2 b c+a d)) \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{4 c d^2}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}+\frac {\left (a b (3 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{4 c d^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-b x^4} \left (c-(b c-a d) x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 c d^2}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}-\frac {\left (a b (3 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{4 c d^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {b c-a d} x^2}{\sqrt {c}}\right ) \sqrt {1-b x^4}} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^2 d^2}-\frac {\left (3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {b c-a d} x^2}{\sqrt {c}}\right ) \sqrt {1-b x^4}} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 c^2 d^2}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}-\frac {3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}-\frac {3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}-\frac {\left (a b (3 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{8 c d^2 \left (a+b x^4\right )^{3/4}}\\ &=\frac {b (3 b c-a d) x \sqrt [4]{a+b x^4}}{4 c d^2}-\frac {(b c-a d) x \left (a+b x^4\right )^{5/4}}{4 c d \left (c+d x^4\right )}-\frac {\sqrt {a} b^{3/2} (3 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 c d^2 \left (a+b x^4\right )^{3/4}}-\frac {3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}-\frac {3 (b c-a d) (2 b c+a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{8 \sqrt [4]{b} c^2 d^2}\\ \end {align*}

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Mathematica [C]  time = 0.59, size = 392, normalized size = 1.11 \[ \frac {2 b x^5 \left (\frac {b x^4}{a}+1\right )^{3/4} \left (a^2 d^2+3 a b c d-3 b^2 c^2\right ) F_1\left (\frac {5}{4};\frac {3}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+\frac {5 c \left (x^5 \left (a+b x^4\right ) \left (a^2 d^2-2 a b c d+b^2 c \left (3 c+2 d x^4\right )\right ) \left (4 a d F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )-5 a c x \left (4 a^3 d^2+a^2 b d^2 x^4+b^3 c x^4 \left (3 c+2 d x^4\right )\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}{\left (c+d x^4\right ) \left (x^4 \left (4 a d F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )-5 a c F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}}{20 c^2 d^2 \left (a+b x^4\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*b*(-3*b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^5*(1 + (b*x^4)/a)^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, -((b*x^4)/a), -((
d*x^4)/c)] + (5*c*(-5*a*c*x*(4*a^3*d^2 + a^2*b*d^2*x^4 + b^3*c*x^4*(3*c + 2*d*x^4))*AppellF1[1/4, 3/4, 1, 5/4,
-((b*x^4)/a), -((d*x^4)/c)] + x^5*(a + b*x^4)*(-2*a*b*c*d + a^2*d^2 + b^2*c*(3*c + 2*d*x^4))*(4*a*d*AppellF1[
5/4, 3/4, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + 3*b*c*AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)]))
)/((c + d*x^4)*(-5*a*c*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + x^4*(4*a*d*AppellF1[5/4, 3/4,
2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + 3*b*c*AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)]))))/(20*c^2
*d^2*(a + b*x^4)^(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((b*x^4+a)^(9/4)/(d*x^4+c)^2,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}}}{\left (d \,x^{4}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((b*x**4+a)**(9/4)/(d*x**4+c)**2,x)

[Out]

Timed out

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