Integrand size = 23, antiderivative size = 95 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {2 \tan (c+d x)}{a d \sqrt {a+a \sec (c+d x)}}+\frac {2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}} \]
2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-2*tan(d*x+c) /a/d/(a+a*sec(d*x+c))^(1/2)+2/3*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)
Time = 3.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.71 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {64 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (\frac {1}{1+\sec (c+d x)}\right )^{7/2} \left (3 \arcsin \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\cos (c+d x)}}}\right ) \cos ^2(c+d x)+\sqrt {\cos (c+d x)} \sqrt {\frac {1}{1+\cos (c+d x)}} (\sin (c+d x)-2 \sin (2 (c+d x)))\right )}{3 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 (a (1+\sec (c+d x)))^{3/2}} \]
(64*Cos[(c + d*x)/2]^6*Cot[(c + d*x)/2]^4*Sec[c + d*x]^5*((1 + Sec[c + d*x ])^(-1))^(7/2)*(3*ArcSin[Tan[(c + d*x)/2]/Sqrt[(1 + Cos[c + d*x])^(-1)]]*C os[c + d*x]^2 + Sqrt[Cos[c + d*x]]*Sqrt[(1 + Cos[c + d*x])^(-1)]*(Sin[c + d*x] - 2*Sin[2*(c + d*x)])))/(3*d*(-1 + Cot[(c + d*x)/2]^2)^2*(a*(1 + Sec[ c + d*x]))^(3/2))
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99, 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.
\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4375 |
\(\displaystyle -\frac {2 a \int \frac {\tan ^4(c+d x)}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {2 a \int \left (\frac {\tan ^2(c+d x)}{a (\sec (c+d x) a+a)}+\frac {1}{a^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}-\frac {1}{a^2}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a \left (-\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2}}+\frac {\tan (c+d x)}{a^2 \sqrt {a \sec (c+d x)+a}}-\frac {\tan ^3(c+d x)}{3 a (a \sec (c+d x)+a)^{3/2}}\right )}{d}\) |
(-2*a*(-(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(5/2)) + Tan[c + d*x]/(a^2*Sqrt[a + a*Sec[c + d*x]]) - Tan[c + d*x]^3/(3*a*(a + a *Sec[c + d*x])^(3/2))))/d
3.2.90.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(173\) vs. \(2(83)=166\).
Time = 3.76 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.83
method | result | size |
default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4 \sin \left (d x +c \right )-\tan \left (d x +c \right )\right )}{3 d \,a^{2} \left (\cos \left (d x +c \right )+1\right )}\) | \(174\) |
-2/3/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(-3*(-cos(d*x+c)/(cos(d *x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c) +1))^(1/2))*cos(d*x+c)-3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d* x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4*sin(d*x+c)-tan(d *x+c))
Time = 0.29 (sec) , antiderivative size = 295, normalized size of antiderivative = 3.11 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (4 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (4 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )\right )}}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \]
[-1/3*(3*(cos(d*x + c)^2 + cos(d*x + c))*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*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(4*cos(d*x + c) - 1)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), -2/3*(3*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arc tan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(4*cos(d*x + c) - 1)*sin( d*x + c))/(a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]
\[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
-1/6*(3*(2*a^2*d*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c os(2*d*x + 2*c) + 1)^(3/4)*(((cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos( 8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*co s(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x + 10*c)* sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 6*sin(6*d*x + 6*c )*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c )^2)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2 *c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d* x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6 *d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(5/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4 *cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2* c) - 4*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c))) - (cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 4*cos(4*d* x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(10*d*x + 10*c)*sin...
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (83) = 166\).
Time = 2.44 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.42 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {-a} {\left (\frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} + \frac {2 \, {\left (\frac {5 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {3 \, \sqrt {2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{3 \, d} \]
-1/3*(3*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^2*sgn(cos(d*x + c))) - lo g(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^2*sgn(cos(d*x + c)))) + 2*(5*sqrt(2)*tan(1/2* d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 3*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)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]