\(\int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\) [700]

Optimal result

Integrand size = 23, antiderivative size = 265 \[ \int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}-\frac {3 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} \left (3 a^2+2 b^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]

[Out]

3/5*(a+b*sec(d*x+c))^(2/3)*tan(d*x+c)/b/d-3/5*a*AppellF1(1/2,-2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d
*x+c))*(a+b*sec(d*x+c))^(2/3)*2^(1/2)*tan(d*x+c)/b^2/d/((a+b*sec(d*x+c))/(a+b))^(2/3)/(1+sec(d*x+c))^(1/2)+1/5
*(3*a^2+2*b^2)*AppellF1(1/2,1/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b*sec(d*x+c))/(a+b))^(1
/3)*2^(1/2)*tan(d*x+c)/b^2/d/(a+b*sec(d*x+c))^(1/3)/(1+sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3925, 4092, 3919, 144, 143} \[ \int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} \left (3 a^2+2 b^2\right ) \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {3 \sqrt {2} a \tan (c+d x) (a+b \sec (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 b d} \]

[In]

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

[Out]

(3*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*b*d) - (3*Sqrt[2]*a*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Sec[c +
d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*b^2*d*Sqrt[1 + Sec[c + d*
x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) + (Sqrt[2]*(3*a^2 + 2*b^2)*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c +
d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(1/3)*Tan[c + d*x])/(5*b^2*d*Sqrt[1 +
Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 3919

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 3925

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-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*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4092

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

Rubi steps \begin{align*} \text {integral}& = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {3 \int \frac {\sec (c+d x) \left (\frac {2 b}{3}-a \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{5 b} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {1}{5} \left (2+\frac {3 a^2}{b^2}\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx-\frac {(3 a) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{5 b^2} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {\left (\left (-2-\frac {3 a^2}{b^2}\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {(3 a \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {\left (3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac {\left (\left (-2-\frac {3 a^2}{b^2}\right ) \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \\ & = \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}-\frac {3 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} \left (2+\frac {3 a^2}{b^2}\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(7195\) vs. \(2(265)=530\).

Time = 44.26 (sec) , antiderivative size = 7195, normalized size of antiderivative = 27.15 \[ \int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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