Integrand size = 23, antiderivative size = 127 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \tan ^3(c+d x)}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac {2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}} \] Output:
-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+2*tan(d*x+c )/a^2/d/(a+a*sec(d*x+c))^(1/2)-2/3*tan(d*x+c)^3/a/d/(a+a*sec(d*x+c))^(3/2) +2/5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)
Leaf count is larger than twice the leaf count of optimal. \(447\) vs. \(2(127)=254\).
Time = 6.05 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.52 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} \cot ^8\left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{9/2} \sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^3 \left (\frac {\sqrt {2} \arcsin \left (\frac {\sqrt {2} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {1-\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {8 \tan ^6\left (\frac {1}{2} (c+d x)\right )}{5 \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^3}+\frac {4 \tan ^4\left (\frac {1}{2} (c+d x)\right )}{3 \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^2}+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (-1+\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )}\right ) \tan ^7(c+d x)}{d (a (1+\sec (c+d x)))^{5/2} \left (1-\frac {2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^{5/2}} \] Input:
Integrate[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(5/2),x]
Output:
(Sqrt[2]*Cot[(c + d*x)/2]^8*((1 + Sec[c + d*x])^(-1))^(9/2)*Sqrt[1 + Tan[( c + d*x)/2]^2]*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3*(( Sqrt[2]*ArcSin[(Sqrt[2]*Tan[(c + d*x)/2])/Sqrt[1 + Tan[(c + d*x)/2]^2]]*Ta n[(c + d*x)/2])/(Sqrt[1 + Tan[(c + d*x)/2]^2]*Sqrt[1 - (2*Tan[(c + d*x)/2] ^2)/(1 + Tan[(c + d*x)/2]^2)]) + (8*Tan[(c + d*x)/2]^6)/(5*(1 + Tan[(c + d *x)/2]^2)^3*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^3) + (4 *Tan[(c + d*x)/2]^4)/(3*(1 + Tan[(c + d*x)/2]^2)^2*(-1 + (2*Tan[(c + d*x)/ 2]^2)/(1 + Tan[(c + d*x)/2]^2))^2) + (2*Tan[(c + d*x)/2]^2)/((1 + Tan[(c + d*x)/2]^2)*(-1 + (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))))*Tan[c + d*x]^7)/(d*(a*(1 + Sec[c + d*x]))^(5/2)*(1 - (2*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2))^(5/2))
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4375, 254, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^6}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a \int \frac {\tan ^6(c+d x)}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {2 a \int \left (\frac {\tan ^4(c+d x)}{a (\sec (c+d x) a+a)^2}-\frac {\tan ^2(c+d x)}{a^2 (\sec (c+d x) a+a)}-\frac {1}{a^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}+\frac {1}{a^3}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{7/2}}-\frac {\tan (c+d x)}{a^3 \sqrt {a \sec (c+d x)+a}}+\frac {\tan ^3(c+d x)}{3 a^2 (a \sec (c+d x)+a)^{3/2}}-\frac {\tan ^5(c+d x)}{5 a (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
Input:
Int[Tan[c + d*x]^6/(a + a*Sec[c + d*x])^(5/2),x]
Output:
(-2*a*(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(7/2) - T an[c + d*x]/(a^3*Sqrt[a + a*Sec[c + d*x]]) + Tan[c + d*x]^3/(3*a^2*(a + a* Sec[c + d*x])^(3/2)) - Tan[c + d*x]^5/(5*a*(a + a*Sec[c + d*x])^(5/2))))/d
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Time = 1.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-23 \sin \left (d x +c \right )+11 \tan \left (d x +c \right )-3 \sec \left (d x +c \right ) \tan \left (d x +c \right )+15 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{15 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )}\) | \(146\) |
Input:
int(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/15/d/a^3*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*(-23*sin(d*x+c)+11*tan (d*x+c)-3*sec(d*x+c)*tan(d*x+c)+15*(1+cos(d*x+c))*(-cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)*arctanh(2^(1/2)/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c )^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.54 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left (15 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (23 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:
integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[-1/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c )^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin( d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(23*cos(d*x + c)^2 - 11*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c ))/(a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2), 2/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (23*cos(d*x + c)^2 - 11*cos(d*x + c ) + 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d* x + c)^3 + a^3*d*cos(d*x + c)^2)]
\[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{6}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**(5/2),x)
Output:
Integral(tan(c + d*x)**6/(a*(sec(c + d*x) + 1))**(5/2), x)
\[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{6}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
-1/30*(20*(9*sin(4*d*x + 4*c) + 16*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 4*(45*cos(4*d*x + 4*c) + 80*cos(2*d* x + 2*c) + 23)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 15*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arc tan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1 /4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arcta n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (cos(2*d*x + 2*c)^2 + s in(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d *x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2* d*x + 2*c) + 1)) - 1) + 2*(a^3*d*cos(2*d*x + 2*c)^2 + a^3*d*sin(2*d*x + 2* c)^2 + 2*a^3*d*cos(2*d*x + 2*c) + a^3*d)*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 + 2*cos(2*d*x + 2*c)^3 + (2*cos(2*d*x + 2*c)^2 + 2*cos (2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + (cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(14* d*x + 14*c) + 6*(cos(2*d*x + 2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(12*d*x + 12*c) + 15*(cos...
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (111) = 222\).
Time = 1.63 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.02 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {-a} {\left (\frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} + \frac {2 \, {\left ({\left (\frac {37 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {40 \, \sqrt {2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {15 \, \sqrt {2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{15 \, d} \] Input:
integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
1/15*(15*sqrt(-a)*(log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/ 2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3)))/(a^3*sgn(cos(d*x + c))) - l og(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a ))^2 + a*(2*sqrt(2) - 3)))/(a^3*sgn(cos(d*x + c)))) + 2*((37*sqrt(2)*tan(1 /2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 40*sqrt(2)/sgn(cos(d*x + c)))*tan(1/ 2*d*x + 1/2*c)^2 + 15*sqrt(2)/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/((a* tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(tan(c + d*x)^6/(a + a/cos(c + d*x))^(5/2),x)
Output:
int(tan(c + d*x)^6/(a + a/cos(c + d*x))^(5/2), x)
\[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{6}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(tan(d*x+c)^6/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**6)/(sec(c + d*x)**3 + 3 *sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3