\(\int \sqrt [3]{a+b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\) [762]

Optimal result

Integrand size = 27, antiderivative size = 27 \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {2} (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {\sqrt {2} a C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+A \text {Int}\left (\sqrt [3]{a+b \sec (c+d x)},x\right ) \]

[Out]

(a+b)*C*AppellF1(1/2,-4/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*(a+b*sec(d*x+c))^(1/3)*2^(1/2)*ta
n(d*x+c)/b/d/((a+b*sec(d*x+c))/(a+b))^(1/3)/(1+sec(d*x+c))^(1/2)-a*C*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))^(1/3)*2^(1/2)*tan(d*x+c)/b/d/((a+b*sec(d*x+c))/(a+b))^(1/3)/(1+se
c(d*x+c))^(1/2)+A*Unintegrable((a+b*sec(d*x+c))^(1/3),x)

Rubi [N/A]

Not integrable

Time = 0.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx \]

[In]

Int[(a + b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[2]*(a + b)*C*AppellF1[1/2, 1/2, -4/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*(a + b*
Sec[c + d*x])^(1/3)*Tan[c + d*x])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) - (Sqrt[2]
*a*C*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])^
(1/3)*Tan[c + d*x])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(1/3)) + A*Defer[Int][(a + b*Se
c[c + d*x])^(1/3), x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sqrt [3]{a+b \sec (c+d x)} (A b-a C \sec (c+d x)) \, dx}{b}+\frac {C \int \sec (c+d x) (a+b \sec (c+d x))^{4/3} \, dx}{b} \\ & = A \int \sqrt [3]{a+b \sec (c+d x)} \, dx-\frac {(a C) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{b}-\frac {(C \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{4/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \\ & = A \int \sqrt [3]{a+b \sec (c+d x)} \, dx+\frac {(a C \tan (c+d x)) \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {\left ((-a-b) C \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{4/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}}} \\ & = \frac {\sqrt {2} (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+A \int \sqrt [3]{a+b \sec (c+d x)} \, dx+\frac {\left (a C \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {a}{-a-b}-\frac {b x}{-a-b}}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}}} \\ & = \frac {\sqrt {2} (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}-\frac {\sqrt {2} a C \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]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+A \int \sqrt [3]{a+b \sec (c+d x)} \, dx \\ \end{align*}

Mathematica [F(-1)]

Timed out. \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {\$Aborted} \]

[In]

Integrate[(a + b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2),x]

[Out]

$Aborted

Maple [N/A] (verified)

Not integrable

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

\[\int \left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]

[In]

int((a+b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x)

[Out]

int((a+b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x)

Fricas [F(-1)]

Timed out. \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 1.88 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt [3]{a + b \sec {\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*sec(d*x+c))**(1/3)*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**(1/3), x)

Maxima [N/A]

Not integrable

Time = 6.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c) + a)^(1/3), x)

Giac [N/A]

Not integrable

Time = 1.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \,d x } \]

[In]

integrate((a+b*sec(d*x+c))^(1/3)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c) + a)^(1/3), x)

Mupad [N/A]

Not integrable

Time = 20.94 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \sqrt [3]{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/3),x)

[Out]

int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/3), x)