\(\int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [323]

Optimal result

Integrand size = 23, antiderivative size = 148 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{5/2}}{5 b^4 d}+\frac {2 (a+b \sec (c+d x))^{7/2}}{7 b^4 d} \] Output:

-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(1/2)/d-2*a*(a^2-2*b^2)*(a+b* 
sec(d*x+c))^(1/2)/b^4/d+2/3*(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(3/2)/b^4/d-6/5 
*a*(a+b*sec(d*x+c))^(5/2)/b^4/d+2/7*(a+b*sec(d*x+c))^(7/2)/b^4/d
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {-\frac {2 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}+\frac {2}{3} \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}-\frac {6}{5} a (a+b \sec (c+d x))^{5/2}+\frac {2}{7} (a+b \sec (c+d x))^{7/2}}{b^4 d} \] Input:

Integrate[Tan[c + d*x]^5/Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

((-2*b^4*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] - 2*a*(a^2 - 2 
*b^2)*Sqrt[a + b*Sec[c + d*x]] + (2*(3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^( 
3/2))/3 - (6*a*(a + b*Sec[c + d*x])^(5/2))/5 + (2*(a + b*Sec[c + d*x])^(7/ 
2))/7)/(b^4*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.83, 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, 1467, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4373

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

\(\Big \downarrow \) 517

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1467

\(\displaystyle -\frac {2 \int \left (-b^6 \sec ^6(c+d x)+3 a b^4 \sec ^4(c+d x)-b^2 \left (3 a^2-2 b^2\right ) \sec ^2(c+d x)+a^3-2 a b^2+\frac {b^4}{a-b^2 \sec ^2(c+d x)}\right )d\sqrt {a+b \sec (c+d x)}}{b^4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (-a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}+\frac {1}{3} b^3 \left (3 a^2-2 b^2\right ) \sec ^3(c+d x)-\frac {b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {3}{5} a b^5 \sec ^5(c+d x)+\frac {1}{7} b^7 \sec ^7(c+d x)\right )}{b^4 d}\)

Input:

Int[Tan[c + d*x]^5/Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(2*(-((b^4*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a]) + (b^3*(3*a 
^2 - 2*b^2)*Sec[c + d*x]^3)/3 - (3*a*b^5*Sec[c + d*x]^5)/5 + (b^7*Sec[c + 
d*x]^7)/7 - a*(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 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 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. \(382\) vs. \(2(128)=256\).

Time = 19.29 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.59

method	result	size
default	\(-\frac {\sqrt {a +b \sec \left (d x +c \right )}\, \left (96 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, a^{4}+48 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, a^{3} b \left (-1-\sec \left (d x +c \right )\right )+4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, a^{2} b^{2} \left (-70-70 \cos \left (d x +c \right )+9 \sec \left (d x +c \right )+9 \sec \left (d x +c \right )^{2}\right )+10 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, a \,b^{3} \left (-3 \sec \left (d x +c \right )^{3}-3 \sec \left (d x +c \right )^{2}+14 \sec \left (d x +c \right )+14\right )+105 \cos \left (d x +c \right ) \sqrt {a}\, \ln \left (4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 a \cos \left (d x +c \right )+2 b \right ) b^{4}\right )}{105 d \,b^{4} a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(383\)

Input:

int(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/105/d/b^4/a*(a+b*sec(d*x+c))^(1/2)/(1+cos(d*x+c))/((b+a*cos(d*x+c))*cos 
(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(96*(1+cos(d*x+c))*((b+a*cos(d*x+c))*cos(d 
*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^4+48*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d 
*x+c))^2)^(1/2)*a^3*b*(-1-sec(d*x+c))+4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+co 
s(d*x+c))^2)^(1/2)*a^2*b^2*(-70-70*cos(d*x+c)+9*sec(d*x+c)+9*sec(d*x+c)^2) 
+10*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a*b^3*(-3*sec(d*x 
+c)^3-3*sec(d*x+c)^2+14*sec(d*x+c)+14)+105*cos(d*x+c)*a^(1/2)*ln(4*((b+a*c 
os(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)
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.57 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\left [\frac {105 \, \sqrt {a} b^{4} \cos \left (d x + c\right )^{3} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (18 \, a^{2} b^{2} \cos \left (d x + c\right ) - 15 \, a b^{3} + 4 \, {\left (12 \, a^{4} - 35 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (12 \, a^{3} b - 35 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{210 \, a b^{4} d \cos \left (d x + c\right )^{3}}, \frac {105 \, \sqrt {-a} b^{4} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{3} - 2 \, {\left (18 \, a^{2} b^{2} \cos \left (d x + c\right ) - 15 \, a b^{3} + 4 \, {\left (12 \, a^{4} - 35 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (12 \, a^{3} b - 35 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{105 \, a b^{4} d \cos \left (d x + c\right )^{3}}\right ] \] Input:

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[1/210*(105*sqrt(a)*b^4*cos(d*x + c)^3*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*c 
os(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*(18*a^2*b^2*cos(d*x + c) - 15*a*b^3 
 + 4*(12*a^4 - 35*a^2*b^2)*cos(d*x + c)^3 - 2*(12*a^3*b - 35*a*b^3)*cos(d* 
x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a*b^4*d*cos(d*x + c)^3 
), 1/105*(105*sqrt(-a)*b^4*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos 
(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))*cos(d*x + c)^3 - 2*(18*a^2 
*b^2*cos(d*x + c) - 15*a*b^3 + 4*(12*a^4 - 35*a^2*b^2)*cos(d*x + c)^3 - 2* 
(12*a^3*b - 35*a*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + 
c)))/(a*b^4*d*cos(d*x + c)^3)]
 

Sympy [F]

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

integrate(tan(d*x+c)**5/(a+b*sec(d*x+c))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.18 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {\frac {105 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {30 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{b^{4}} - \frac {126 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} a}{b^{4}} + \frac {210 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{2}}{b^{4}} - \frac {210 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a^{3}}{b^{4}} - \frac {140 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{b^{2}} + \frac {420 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}{b^{2}}}{105 \, d} \] Input:

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/105*(105*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + 
c)) + sqrt(a)))/sqrt(a) + 30*(a + b/cos(d*x + c))^(7/2)/b^4 - 126*(a + b/c 
os(d*x + c))^(5/2)*a/b^4 + 210*(a + b/cos(d*x + c))^(3/2)*a^2/b^4 - 210*sq 
rt(a + b/cos(d*x + c))*a^3/b^4 - 140*(a + b/cos(d*x + c))^(3/2)/b^2 + 420* 
sqrt(a + b/cos(d*x + c))*a/b^2)/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 699 vs. \(2 (128) = 256\).

Time = 0.67 (sec) , antiderivative size = 699, normalized size of antiderivative = 4.72 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

2/105*(105*arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/ 
2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + 
 a + b) + sqrt(a - b))/sqrt(-a))/sqrt(-a) - 2*(105*(sqrt(a - b)*tan(1/2*d* 
x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 
2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^6 - 840*(sqrt(a - b)*tan(1/2*d*x + 1/ 
2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*ta 
n(1/2*d*x + 1/2*c)^2 + a + b))^5*sqrt(a - b) + 35*(sqrt(a - b)*tan(1/2*d*x 
 + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2 
*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^4*(27*a - 23*b) + 280*(sqrt(a - b)*tan 
(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2* 
c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^3*(3*a + 4*b)*sqrt(a - b) - 21 
*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*t 
an(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^2*(65*a^2 - 2 
*a*b - 15*b^2) + 315*a^3 + 707*a^2*b - 7*a*b^2 - 55*b^3 - 56*(sqrt(a - b)* 
tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1 
/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))*(19*a*b + 5*b^2)*sqrt(a - b 
))/(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b 
*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) - sqrt(a - b 
))^7)/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(1/2),x)
 

Output:

int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(1/2), x)
 

Reduce [F]

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

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

Output:

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