\(\int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [332]

Optimal result

Integrand size = 23, antiderivative size = 148 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d} \] Output:

-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2*(a^2-b^2)^2/a/b^4/d 
/(a+b*sec(d*x+c))^(1/2)+2*(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(1/2)/b^4/d-2*a*( 
a+b*sec(d*x+c))^(3/2)/b^4/d+2/5*(a+b*sec(d*x+c))^(5/2)/b^4/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.55 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (5 b^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )+a \left (4 a \left (4 a^2-5 b^2\right )+2 b \left (4 a^2-5 b^2\right ) \sec (c+d x)-2 a b^2 \sec ^2(c+d x)+b^3 \sec ^3(c+d x)\right )\right )}{5 a b^4 d \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(2*(5*b^4*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sec[c + d*x])/a] + a*(4*a 
*(4*a^2 - 5*b^2) + 2*b*(4*a^2 - 5*b^2)*Sec[c + d*x] - 2*a*b^2*Sec[c + d*x] 
^2 + b^3*Sec[c + d*x]^3)))/(5*a*b^4*d*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4373, 517, 25, 1584, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}{\left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4373

\(\displaystyle \frac {\int \frac {\cos (c+d x) \left (b^2-b^2 \sec ^2(c+d x)\right )^2}{b (a+b \sec (c+d x))^{3/2}}d(b \sec (c+d x))}{b^4 d}\)

\(\Big \downarrow \) 517

\(\displaystyle \frac {2 \int -\frac {\cos ^2(c+d x) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}{b^2 \left (a-b^2 \sec ^2(c+d x)\right )}d\sqrt {a+b \sec (c+d x)}}{b^4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 \int \frac {\cos ^2(c+d x) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}{b^2 \left (a-b^2 \sec ^2(c+d x)\right )}d\sqrt {a+b \sec (c+d x)}}{b^4 d}\)

\(\Big \downarrow \) 1584

\(\displaystyle -\frac {2 \int \left (-\sec ^4(c+d x) b^4+\frac {b^4}{a \left (a-b^2 \sec ^2(c+d x)\right )}+3 a \sec ^2(c+d x) b^2-3 a^2 \left (1-\frac {2 b^2}{3 a^2}\right )+\frac {\left (a^2-b^2\right )^2 \cos ^2(c+d x)}{a b^2}\right )d\sqrt {a+b \sec (c+d x)}}{b^4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-\frac {b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a b}+\left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}-a b^3 \sec ^3(c+d x)+\frac {1}{5} b^5 \sec ^5(c+d x)\right )}{b^4 d}\)

Input:

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

Output:

(2*(-((b^4*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/a^(3/2)) + ((a^2 - b 
^2)^2*Cos[c + d*x])/(a*b) - a*b^3*Sec[c + d*x]^3 + (b^5*Sec[c + d*x]^5)/5 
+ (3*a^2 - 2*b^2)*Sqrt[a + b*Sec[c + d*x]]))/(b^4*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 517

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1))   Subst[Int[x^(2*n + 1)*(-c + x^ 
2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

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 4373

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1))   Subst[Int[(b^2 - x^ 
2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, 
 d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(765\) vs. \(2(132)=264\).

Time = 14.67 (sec) , antiderivative size = 766, normalized size of antiderivative = 5.18

method	result	size
default	\(\text {Expression too large to display}\)	\(766\)

Input:

int(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/5/d/a^2/b^4*(a+b*sec(d*x+c))^(1/2)/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d 
*x+c))^2)^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)^2*a^2+2*a*b*cos(d*x+c)+b^2)*(32 
*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^6*(cos(d*x+c)^3+co 
s(d*x+c)^2)+48*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^5*b* 
(cos(d*x+c)+cos(d*x+c)^2)+4*(3-10*cos(d*x+c)^3-10*cos(d*x+c)^2+3*cos(d*x+c 
))*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^4*b^2+2*((b+a*co 
s(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^3*b^3*(-30*cos(d*x+c)^2-30* 
cos(d*x+c)-sec(d*x+c)-1)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^ 
(1/2)*a^2*b^4*(5*cos(d*x+c)^3+5*cos(d*x+c)^2+sec(d*x+c)^2-10*cos(d*x+c)+se 
c(d*x+c)-10)+10*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a*b^5 
*(cos(d*x+c)+cos(d*x+c)^2)-5*a^(5/2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+ 
cos(d*x+c))^2)^(1/2)*a^(1/2)*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d* 
x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*b^4*cos(d*x+c)^3-10*a^(3/ 
2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^(1/2)*cos(d 
*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*c 
os(d*x+c)+2*b)*b^5*cos(d*x+c)^2-5*a^(1/2)*ln(4*((b+a*cos(d*x+c))*cos(d*x+c 
)/(1+cos(d*x+c))^2)^(1/2)*a^(1/2)*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*c 
os(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*b^6*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 467, normalized size of antiderivative = 3.16 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\left [\frac {5 \, {\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} - {\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{10 \, {\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}, \frac {5 \, {\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) - 2 \, {\left (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} - {\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{5 \, {\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/10*(5*(a*b^4*cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*sqrt(a)*log(-8*a^2*co 
s(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 + 4*(2*a*cos(d*x + c)^2 + b*cos(d* 
x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) - 4*(2*a^3*b^2*co 
s(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3 - 2* 
(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + 
c)))/(a^3*b^4*d*cos(d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2), 1/5*(5*(a*b^4* 
cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*co 
s(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - 2*(2* 
a^3*b^2*cos(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + 
 c)^3 - 2*(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/ 
cos(d*x + c)))/(a^3*b^4*d*cos(d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {10}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a} + \frac {2 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{4}} - \frac {10 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{4}} + \frac {30 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a^{2}}{b^{4}} + \frac {10 \, a^{3}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{4}} - \frac {20 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}}}{b^{2}} - \frac {20 \, a}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{2}}}{5 \, d} \] Input:

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

Output:

1/5*(5*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) 
+ sqrt(a)))/a^(3/2) + 10/(sqrt(a + b/cos(d*x + c))*a) + 2*(a + b/cos(d*x + 
 c))^(5/2)/b^4 - 10*(a + b/cos(d*x + c))^(3/2)*a/b^4 + 30*sqrt(a + b/cos(d 
*x + c))*a^2/b^4 + 10*a^3/(sqrt(a + b/cos(d*x + c))*b^4) - 20*sqrt(a + b/c 
os(d*x + c))/b^2 - 20*a/(sqrt(a + b/cos(d*x + c))*b^2))/d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x)*b + a)*tan(c + d*x)**5)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)