\(\int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\) [153]

Optimal result

Integrand size = 23, antiderivative size = 371 \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {37\ 3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]

[Out]

99/80*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)+3/8*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)-3/40*(a+a*sec(d
*x+c))^(2/3)*tan(d*x+c)/a/d+37/160*3^(3/4)*((2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c
))^(1/3)*(1+3^(1/2)))^2)^(1/2)/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))*(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(
1/2)))*EllipticF((1-(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1-3^(1/2)))^2/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)
^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(2^(1/3)-(1+sec(d*x+c))^(1/3))*((2^(2/3)+2^(1/3)*(1+sec(d*x+c))^(1/3)+(1+sec(d
*x+c))^(2/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)*2^(2/3)/d/(1-sec(d*x+c))/(a+a*sec
(d*x+c))^(1/3)/(-(1+sec(d*x+c))^(1/3)*(2^(1/3)-(1+sec(d*x+c))^(1/3))/(2^(1/3)-(1+sec(d*x+c))^(1/3)*(1+3^(1/2))
)^2)^(1/2)

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3909, 4095, 4086, 3913, 3912, 65, 231} \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {37\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac {3 \tan (c+d x) \sec ^2(c+d x)}{8 d \sqrt [3]{a \sec (c+d x)+a}}-\frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{40 a d}+\frac {99 \tan (c+d x)}{80 d \sqrt [3]{a \sec (c+d x)+a}} \]

[In]

Int[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^(1/3),x]

[Out]

(99*Tan[c + d*x])/(80*d*(a + a*Sec[c + d*x])^(1/3)) + (3*Sec[c + d*x]^2*Tan[c + d*x])/(8*d*(a + a*Sec[c + d*x]
)^(1/3)) - (3*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(40*a*d) + (37*3^(3/4)*EllipticF[ArcCos[(2^(1/3) - (1 -
Sqrt[3])*(1 + Sec[c + d*x])^(1/3))/(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))], (2 + Sqrt[3])/4]*(2^(
1/3) - (1 + Sec[c + d*x])^(1/3))*Sqrt[(2^(2/3) + 2^(1/3)*(1 + Sec[c + d*x])^(1/3) + (1 + Sec[c + d*x])^(2/3))/
(2^(1/3) - (1 + Sqrt[3])*(1 + Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(80*2^(1/3)*d*(1 - Sec[c + d*x])*(a + a*Se
c[c + d*x])^(1/3)*Sqrt[-(((1 + Sec[c + d*x])^(1/3)*(2^(1/3) - (1 + Sec[c + d*x])^(1/3)))/(2^(1/3) - (1 + Sqrt[
3])*(1 + Sec[c + d*x])^(1/3))^2)])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 3909

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d^2)
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 2)/(f*(m + n - 1))), x] + Dist[d^2/(b*(m + n - 1))
, Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*m*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, m}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 2] && NeQ[m + n - 1, 0] && IntegerQ[n]

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^2*d
*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])), Subst[Int[(d*x)^(n - 1)*((a + b*x)^(m -
1/2)/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 3913

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[a^Int
Part[m]*((a + b*Csc[e + f*x])^FracPart[m]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]), Int[(1 + (b/a)*Csc[e + f*x])^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 4086

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*B*m + A*b*(m + 1))/(b
*(m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B
, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4095

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \int \frac {\sec ^2(c+d x) \left (2 a-\frac {1}{3} a \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{8 a} \\ & = \frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {9 \int \frac {\sec (c+d x) \left (-\frac {2 a^2}{9}+\frac {11}{3} a^2 \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{40 a^2} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}-\frac {37}{80} \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}-\frac {\left (37 \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{80 \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {(37 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{80 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {(111 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{40 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \\ & = \frac {99 \tan (c+d x)}{80 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 \sec ^2(c+d x) \tan (c+d x)}{8 d \sqrt [3]{a+a \sec (c+d x)}}-\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{40 a d}+\frac {37\ 3^{3/4} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{80 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.40 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.42 \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {\left (-4 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )+16 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )-7 \sqrt [6]{2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right )+3 \sec ^2(c+d x) \sqrt [6]{1+\sec (c+d x)}\right ) \tan (c+d x)}{8 d \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a (1+\sec (c+d x))}} \]

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^(1/3),x]

[Out]

((-4*2^(1/6)*Hypergeometric2F1[-7/6, 1/2, 3/2, (1 - Sec[c + d*x])/2] + 16*2^(1/6)*Hypergeometric2F1[-1/6, 1/2,
3/2, (1 - Sec[c + d*x])/2] - 7*2^(1/6)*Hypergeometric2F1[1/2, 5/6, 3/2, (1 - Sec[c + d*x])/2] + 3*Sec[c + d*x
]^2*(1 + Sec[c + d*x])^(1/6))*Tan[c + d*x])/(8*d*(1 + Sec[c + d*x])^(1/6)*(a*(1 + Sec[c + d*x]))^(1/3))

Maple [F]

\[\int \frac {\sec \left (d x +c \right )^{4}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]

[In]

int(sec(d*x+c)^4/(a+a*sec(d*x+c))^(1/3),x)

[Out]

int(sec(d*x+c)^4/(a+a*sec(d*x+c))^(1/3),x)

Fricas [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral(sec(d*x + c)^4/(a*sec(d*x + c) + a)^(1/3), x)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate(sec(d*x+c)**4/(a+a*sec(d*x+c))**(1/3),x)

[Out]

Integral(sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(1/3), x)

Maxima [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(a*sec(d*x + c) + a)^(1/3), x)

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(a*sec(d*x + c) + a)^(1/3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]

[In]

int(1/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(1/3)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(1/3)), x)