3.198 \(\int \frac {1}{(a+b x^4)^{13/4} (c+d x^4)} \, dx\)

Optimal. Leaf size=233 \[ \frac {b x (8 b c-17 a d)}{45 a^2 \left (a+b x^4\right )^{5/4} (b c-a d)^2}+\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{45 a^3 \sqrt [4]{a+b x^4} (b c-a d)^3}-\frac {d^3 \tan ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \tanh ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)} \]

[Out]

1/9*b*x/a/(-a*d+b*c)/(b*x^4+a)^(9/4)+1/45*b*(-17*a*d+8*b*c)*x/a^2/(-a*d+b*c)^2/(b*x^4+a)^(5/4)+1/45*b*(113*a^2
*d^2-100*a*b*c*d+32*b^2*c^2)*x/a^3/(-a*d+b*c)^3/(b*x^4+a)^(1/4)-1/2*d^3*arctan((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x
^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(13/4)-1/2*d^3*arctanh((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/(-a
*d+b*c)^(13/4)

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Rubi [A]  time = 0.29, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {414, 527, 12, 377, 212, 208, 205} \[ \frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{45 a^3 \sqrt [4]{a+b x^4} (b c-a d)^3}+\frac {b x (8 b c-17 a d)}{45 a^2 \left (a+b x^4\right )^{5/4} (b c-a d)^2}-\frac {d^3 \tan ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \tanh ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]

[Out]

(b*x)/(9*a*(b*c - a*d)*(a + b*x^4)^(9/4)) + (b*(8*b*c - 17*a*d)*x)/(45*a^2*(b*c - a*d)^2*(a + b*x^4)^(5/4)) +
(b*(32*b^2*c^2 - 100*a*b*c*d + 113*a^2*d^2)*x)/(45*a^3*(b*c - a*d)^3*(a + b*x^4)^(1/4)) - (d^3*ArcTan[((b*c -
a*d)^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4))])/(2*c^(3/4)*(b*c - a*d)^(13/4)) - (d^3*ArcTanh[((b*c - a*d)^(1/4)*x
)/(c^(1/4)*(a + b*x^4)^(1/4))])/(2*c^(3/4)*(b*c - a*d)^(13/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}-\frac {\int \frac {-8 b c+9 a d-8 b d x^4}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx}{9 a (b c-a d)}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {\int \frac {32 b^2 c^2-68 a b c d+45 a^2 d^2+4 b d (8 b c-17 a d) x^4}{\left (a+b x^4\right )^{5/4} \left (c+d x^4\right )} \, dx}{45 a^2 (b c-a d)^2}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {\int \frac {45 a^3 d^3}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{45 a^3 (b c-a d)^3}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \int \frac {1}{\sqrt [4]{a+b x^4} \left (c+d x^4\right )} \, dx}{(b c-a d)^3}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{c-(b c-a d) x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{(b c-a d)^3}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {c}-\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt {c} (b c-a d)^3}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {c}+\sqrt {b c-a d} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt {c} (b c-a d)^3}\\ &=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \tan ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 5.44, size = 231, normalized size = 0.99 \[ \frac {b x \left (\left (a+b x^4\right )^2 \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )+5 a^2 (b c-a d)^2+a \left (a+b x^4\right ) (a d-b c) (17 a d-8 b c)\right )}{45 a^3 \left (a+b x^4\right )^{9/4} (b c-a d)^3}-\frac {d^3 \left (-\log \left (\sqrt [4]{c}-\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{a x^4+b}}\right )+\log \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{a x^4+b}}+\sqrt [4]{c}\right )+2 \tan ^{-1}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a x^4+b}}\right )\right )}{4 c^{3/4} (b c-a d)^{13/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]

[Out]

(b*x*(5*a^2*(b*c - a*d)^2 + a*(-(b*c) + a*d)*(-8*b*c + 17*a*d)*(a + b*x^4) + (32*b^2*c^2 - 100*a*b*c*d + 113*a
^2*d^2)*(a + b*x^4)^2))/(45*a^3*(b*c - a*d)^3*(a + b*x^4)^(9/4)) - (d^3*(2*ArcTan[((b*c - a*d)^(1/4)*x)/(c^(1/
4)*(b + a*x^4)^(1/4))] - Log[c^(1/4) - ((b*c - a*d)^(1/4)*x)/(b + a*x^4)^(1/4)] + Log[c^(1/4) + ((b*c - a*d)^(
1/4)*x)/(b + a*x^4)^(1/4)]))/(4*c^(3/4)*(b*c - a*d)^(13/4))

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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^4+a)^(13/4)/(d*x^4+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {13}{4}} \left (d \,x^{4}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^4+a)^(13/4)/(d*x^4+c),x)

[Out]

int(1/(b*x^4+a)^(13/4)/(d*x^4+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,x^4+a\right )}^{13/4}\,\left (d\,x^4+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^4)^(13/4)*(c + d*x^4)),x)

[Out]

int(1/((a + b*x^4)^(13/4)*(c + d*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{4}\right )^{\frac {13}{4}} \left (c + d x^{4}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**4+a)**(13/4)/(d*x**4+c),x)

[Out]

Integral(1/((a + b*x**4)**(13/4)*(c + d*x**4)), x)

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