\(\int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\) [763]

Optimal result

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

[Out]

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

Rubi [N/A]

Not integrable

Time = 0.39 (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 = {} \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx \]

[In]

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

[Out]

(Sqrt[2]*C*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])/(b*d*Sqrt[1 + Sec[c + d*x]]*((a + b*Sec[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*a*C*App
ellF1[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])/(b*d*Sqrt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3)) + A*Defer[Int][(a + b*Sec[c + d*x
])^(-1/3), x]

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

Mathematica [F(-1)]

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

[In]

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

[Out]

$Aborted

Maple [N/A] (verified)

Not integrable

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

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 0.87 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{a + b \sec {\left (c + d x \right )}}}\, dx \]

[In]

integrate((A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(1/3),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 = 11.37 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/3),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 = 1.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/3),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 = 19.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]

[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)