\(\int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx\) [644]

Optimal result

Integrand size = 25, antiderivative size = 687 \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=-\frac {a \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{2 \sqrt {3} b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {a \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{2 \sqrt {3} b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {a \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{3 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {a \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {a \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {1}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )} \] Output:

1/6*a*arctan(-1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)*3^(1/2)/(a^2+b^ 
2)^(1/6))*(d*sec(f*x+e))^(5/3)*3^(1/2)/b^(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+ 
e)^2)^(5/6)+1/6*a*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)*3^(1 
/2)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(5/3)*3^(1/2)/b^(2/3)/(a^2+b^2)^(7/6)/ 
f/(sec(f*x+e)^2)^(5/6)-1/3*a*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2 
)^(1/6))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+e)^2)^(5/ 
6)+1/12*a*ln((a^2+b^2)^(1/3)-b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+ 
b^(2/3)*(sec(f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6) 
/f/(sec(f*x+e)^2)^(5/6)-1/12*a*ln((a^2+b^2)^(1/3)+b^(1/3)*(a^2+b^2)^(1/6)* 
(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(5/3)/b^ 
(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+e)^2)^(5/6)+AppellF1(1/2,2,1/6,3/2,b^2*ta 
n(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(5/3)*tan(f*x+e)/a^2/f/(sec(f 
*x+e)^2)^(5/6)+1/3*b^2*AppellF1(3/2,2,1/6,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f* 
x+e)^2)*(d*sec(f*x+e))^(5/3)*tan(f*x+e)^3/a^4/f/(sec(f*x+e)^2)^(5/6)-a*b*( 
d*sec(f*x+e))^(5/3)/(a^2+b^2)/f/(a^2-b^2*tan(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 118.43 (sec) , antiderivative size = 8003, normalized size of antiderivative = 11.65 \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x])^2,x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 545, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3994, 505, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int \frac {1}{(a+b \tan (e+f x))^2 \sqrt [6]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{5/6}}\)

\(\Big \downarrow \) 505

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int \left (\frac {a^2}{\sqrt [6]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 b \tan (e+f x) a}{\sqrt [6]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x)}{\sqrt [6]{\tan ^2(e+f x)+1} \left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{5/6}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}-\frac {a \sqrt [3]{b} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{7/6}}+\frac {a \sqrt [3]{b} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{7/6}}-\frac {a \sqrt [3]{b} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt [6]{a^2+b^2}}\right )}{3 \left (a^2+b^2\right )^{7/6}}-\frac {a b^2 \left (\tan ^2(e+f x)+1\right )^{5/6}}{\left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {a \sqrt [3]{b} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{7/6}}-\frac {a \sqrt [3]{b} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{7/6}}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {1}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f \sec ^2(e+f x)^{5/6}}\)

Input:

Int[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Sec[e + f*x])^(5/3)*(-1/2*(a*b^(1/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(1 
+ Tan[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(Sqrt[3]*(a^2 + b^2 
)^(7/6)) + (a*b^(1/3)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(1 + Tan[e + f*x]^2)^( 
1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(2*Sqrt[3]*(a^2 + b^2)^(7/6)) - (a*b^( 
1/3)*ArcTanh[(b^(1/3)*(1 + Tan[e + f*x]^2)^(1/6))/(a^2 + b^2)^(1/6)])/(3*( 
a^2 + b^2)^(7/6)) + (a*b^(1/3)*Log[(a^2 + b^2)^(1/3) - b^(1/3)*(a^2 + b^2) 
^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f*x]^2)^(1/3)])/( 
12*(a^2 + b^2)^(7/6)) - (a*b^(1/3)*Log[(a^2 + b^2)^(1/3) + b^(1/3)*(a^2 + 
b^2)^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f*x]^2)^(1/3) 
])/(12*(a^2 + b^2)^(7/6)) + (b*AppellF1[1/2, 2, 1/6, 3/2, (b^2*Tan[e + f*x 
]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a^2 + (b^3*AppellF1[3/2, 2, 1/6, 
5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x]^3)/(3*a^4) - 
(a*b^2*(1 + Tan[e + f*x]^2)^(5/6))/((a^2 + b^2)*(a^2 - b^2*Tan[e + f*x]^2) 
)))/(b*f*(Sec[e + f*x]^2)^(5/6))
 

Defintions of rubi rules used

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( 
-n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]))   Subst[Int[(a + x)^n*(1 + 
x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

Maple [F]

\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]

Input:

int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x)
 

Output:

int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate((d*sec(f*x+e))**(5/3)/(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((d*sec(e + f*x))**(5/3)/(a + b*tan(e + f*x))**2, x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/3)/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x))^2,x)
 

Output:

int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx=\frac {d^{\frac {5}{3}} \left (-3 \sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )+5 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) b^{2} f +5 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a b f +8 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +8 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f +15 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +15 \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f \right )}{12 a f \left (\tan \left (f x +e \right ) b +a \right )} \] Input:

int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x)
 

Output:

(d**(2/3)*d*( - 3*sec(e + f*x)**(2/3)*sec(e + f*x)*tan(e + f*x) + 5*int((s 
ec(e + f*x)**(2/3)*sec(e + f*x)*tan(e + f*x)**3)/(tan(e + f*x)**2*b**2 + 2 
*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*b**2*f + 5*int((sec(e + f*x)**(2 
/3)*sec(e + f*x)*tan(e + f*x)**3)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a 
*b + a**2),x)*a*b*f + 8*int((sec(e + f*x)**(2/3)*sec(e + f*x)*tan(e + f*x) 
**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*a* 
b*f + 8*int((sec(e + f*x)**(2/3)*sec(e + f*x)*tan(e + f*x)**2)/(tan(e + f* 
x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*f + 15*int((sec(e + f*x)** 
(2/3)*sec(e + f*x))/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)* 
tan(e + f*x)*a*b*f + 15*int((sec(e + f*x)**(2/3)*sec(e + f*x))/(tan(e + f* 
x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*f))/(12*a*f*(tan(e + f*x)* 
b + a))