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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.294959, antiderivative size = 233, normalized size of antiderivative = 1., 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 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 12

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

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 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 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 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]

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*}

Mathematica [C]  time = 4.36266, size = 1172, normalized size = 5.03 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-16575*c^5*(b*c - a*d)^2*x^8*(a + b*x^4)^2 - 39780*c^4*d*(b*c - a*d)^2*x^12*(a + b*x^4)^2 - 35360*c^3*d^2*(b
*c - a*d)^2*x^16*(a + b*x^4)^2 - 10880*c^2*d^3*(b*c - a*d)^2*x^20*(a + b*x^4)^2 - 29835*c^6*(b*c - a*d)*x^4*(a
 + b*x^4)^3 - 71604*c^5*d*(b*c - a*d)*x^8*(a + b*x^4)^3 - 63648*c^4*d^2*(b*c - a*d)*x^12*(a + b*x^4)^3 - 19584
*c^3*d^3*(b*c - a*d)*x^16*(a + b*x^4)^3 - 149175*c^7*(a + b*x^4)^4 - 358020*c^6*d*x^4*(a + b*x^4)^4 - 318240*c
^5*d^2*x^8*(a + b*x^4)^4 - 97920*c^4*d^3*x^12*(a + b*x^4)^4 + 149175*c^7*(a + b*x^4)^4*Hypergeometric2F1[1/4,
1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 358020*c^6*d*x^4*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b
*c - a*d)*x^4)/(c*(a + b*x^4))] + 318240*c^5*d^2*x^8*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)
*x^4)/(c*(a + b*x^4))] + 97920*c^4*d^3*x^12*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*
(a + b*x^4))] + 13620*c^3*(b*c - a*d)^4*x^16*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4)
)] + 36900*c^2*d*(b*c - a*d)^4*x^20*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 3384
0*c*d^2*(b*c - a*d)^4*x^24*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 10560*d^3*(b*
c - a*d)^4*x^28*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 6480*c^3*(b*c - a*d)^4*x
^16*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 18720*c^2*d*(b*c - a*d)^4*
x^20*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 18000*c*d^2*(b*c - a*d)^4
*x^24*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 5760*d^3*(b*c - a*d)^4*x
^28*HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 960*c^3*(b*c - a*d)^4*x^16
*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 2880*c^2*d*(b*c - a*d)^
4*x^20*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 2880*c*d^2*(b*c -
 a*d)^4*x^24*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 960*d^3*(b*
c - a*d)^4*x^28*HypergeometricPFQ[{2, 2, 2, 17/4}, {1, 1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))])/(11475*c^
5*(-(b*c) + a*d)^3*x^11*(a + b*x^4)^(17/4))

________________________________________________________________________________________

Maple [F]  time = 0.425, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{4}+c} \left ( b{x}^{4}+a \right ) ^{-{\frac{13}{4}}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{13}{4}}{\left (d x^{4} + c\right )}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{13}{4}}{\left (d x^{4} + c\right )}}\,{d x} \end{align*}

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)