3.2.99 \(\int \frac {(a+b x^4)^{9/4}}{c+d x^4} \, dx\) [199]

Optimal. Leaf size=316 \[ -\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {\sqrt {a} b^{3/2} (6 b c-11 a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 d^2 \left (a+b x^4\right )^{3/4}}+\frac {(b c-a d)^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{2 \sqrt [4]{b} c d^2}+\frac {(b c-a d)^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\right |-1\right )}{2 \sqrt [4]{b} c d^2} \]

[Out]

-1/12*b*(-11*a*d+6*b*c)*x*(b*x^4+a)^(1/4)/d^2+1/6*b*x*(b*x^4+a)^(5/4)/d+1/12*b^(3/2)*(-11*a*d+6*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)/d^2/(b*x^4+a)^(3/4)+1/2*(-a*d+b*c)^2*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/d^2+1/2*(-a*
d+b*c)^2*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/d^2

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Rubi [A]
time = 0.21, antiderivative size = 316, 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 = {427, 542, 543, 243, 342, 281, 237, 416, 418, 1232} \begin {gather*} \frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d)^2 \Pi \left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c d^2}+\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d)^2 \Pi \left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right )\right |-1\right )}{2 \sqrt [4]{b} c d^2}+\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (6 b c-11 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 d^2 \left (a+b x^4\right )^{3/4}}-\frac {b x \sqrt [4]{a+b x^4} (6 b c-11 a d)}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(9/4)/(c + d*x^4),x]

[Out]

-1/12*(b*(6*b*c - 11*a*d)*x*(a + b*x^4)^(1/4))/d^2 + (b*x*(a + b*x^4)^(5/4))/(6*d) + (Sqrt[a]*b^(3/2)*(6*b*c -
11*a*d)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*d^2*(a + b*x^4)^(3/4)) +
((b*c - a*d)^2*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])/(2*b^(1/4)*c*d^2) + ((b*c - a*d)^2*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4]*Elli
pticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c]), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1])/(2*b^(1/4)*c*d^2)

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 243

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 281

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 342

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 416

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 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 427

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

Rule 542

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 543

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 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 {\left (a+b x^4\right )^{9/4}}{c+d x^4} \, dx &=\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {\int \frac {\sqrt [4]{a+b x^4} \left (-a (b c-6 a d)-b (6 b c-11 a d) x^4\right )}{c+d x^4} \, dx}{6 d}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {\int \frac {a \left (6 b^2 c^2-13 a b c d+12 a^2 d^2\right )+b \left (12 b^2 c^2-30 a b c d+23 a^2 d^2\right ) x^4}{\left (a+b x^4\right )^{3/4} \left (c+d x^4\right )} \, dx}{12 d^2}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}-\frac {(a b (6 b c-11 a d)) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{12 d^2}+\frac {(b c-a d)^2 \int \frac {\sqrt [4]{a+b x^4}}{c+d x^4} \, dx}{d^2}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}-\frac {\left (a b (6 b c-11 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}{12 d^2 \left (a+b x^4\right )^{3/4}}+\frac {\left ((b c-a d)^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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 )}{d^2}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {\left (a b (6 b c-11 a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{12 d^2 \left (a+b x^4\right )^{3/4}}+\frac {\left ((b c-a d)^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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 )}{2 c d^2}+\frac {\left ((b c-a d)^2 \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4}\right ) \text {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 )}{2 c d^2}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {(b c-a d)^2 \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 )}{2 \sqrt [4]{b} c d^2}+\frac {(b c-a d)^2 \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 )}{2 \sqrt [4]{b} c d^2}+\frac {\left (a b (6 b c-11 a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{24 d^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac {b (6 b c-11 a d) x \sqrt [4]{a+b x^4}}{12 d^2}+\frac {b x \left (a+b x^4\right )^{5/4}}{6 d}+\frac {\sqrt {a} b^{3/2} (6 b c-11 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 )}{12 d^2 \left (a+b x^4\right )^{3/4}}+\frac {(b c-a d)^2 \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 )}{2 \sqrt [4]{b} c d^2}+\frac {(b c-a d)^2 \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 )}{2 \sqrt [4]{b} c d^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.47, size = 294, normalized size = 0.93 \begin {gather*} \frac {x \left (5 b \left (a+b x^4\right ) \left (-6 b c+13 a d+2 b d x^4\right )+\frac {b \left (12 b^2 c^2-30 a b c d+23 a^2 d^2\right ) x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} F_1\left (\frac {5}{4};\frac {3}{4},1;\frac {9}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c}-\frac {25 a^2 c \left (6 b^2 c^2-13 a b c d+12 a^2 d^2\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 )}{\left (c+d x^4\right ) \left (-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 )+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 )\right )}\right )}{60 d^2 \left (a+b x^4\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(5*b*(a + b*x^4)*(-6*b*c + 13*a*d + 2*b*d*x^4) + (b*(12*b^2*c^2 - 30*a*b*c*d + 23*a^2*d^2)*x^4*(1 + (b*x^4)
/a)^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])/c - (25*a^2*c*(6*b^2*c^2 - 13*a*b*c*d + 12*a
^2*d^2)*AppellF1[1/4, 3/4, 1, 5/4, -((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*Appe
llF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))))/(60*d^2*(a + b*x^4)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}}}{d \,x^{4}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^4 + a)^(9/4)/(d*x^4 + 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((b*x^4+a)^(9/4)/(d*x^4+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 {\left (a + b x^{4}\right )^{\frac {9}{4}}}{c + d x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(9/4)/(c + d*x^4),x)

[Out]

int((a + b*x^4)^(9/4)/(c + d*x^4), x)

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