Integrand size = 23, antiderivative size = 160 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {14 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^5 \tan ^5(c+d x)}{d (a+a \sec (c+d x))^{5/2}}+\frac {2 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}} \]
-2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2*a^3*tan(d *x+c)/d/(a+a*sec(d*x+c))^(1/2)+14/3*a^4*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3 /2)+2*a^5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+2/7*a^6*tan(d*x+c)^7/d/(a+ a*sec(d*x+c))^(7/2)
Time = 6.42 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=-\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (1+\sec (c+d x))} \left (42 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)-35 \sin \left (\frac {1}{2} (c+d x)\right )+7 \sin \left (\frac {3}{2} (c+d x)\right )-21 \sin \left (\frac {5}{2} (c+d x)\right )+5 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{42 d} \]
-1/42*(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*(42* Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(7/2) - 35*Sin[(c + d*x)/2] + 7*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] + 5*Sin[(7*(c + d*x))/2]))/d
Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4375, 364, 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.
| \(\displaystyle \int \tan ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a^4 \int \frac {\tan ^2(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^3}{(\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle -\frac {2 a^4 \int \left (\frac {a^2 \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {5 a \tan ^4(c+d x)}{(\sec (c+d x) a+a)^2}+\frac {7 \tan ^2(c+d x)}{\sec (c+d x) a+a}+\frac {1}{a}-\frac {1}{a \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^4 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2}}-\frac {a^2 \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {a \tan ^5(c+d x)}{(a \sec (c+d x)+a)^{5/2}}-\frac {7 \tan ^3(c+d x)}{3 (a \sec (c+d x)+a)^{3/2}}-\frac {\tan (c+d x)}{a \sqrt {a \sec (c+d x)+a}}\right )}{d}\) |
(-2*a^4*(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(3/2) - Tan[c + d*x]/(a*Sqrt[a + a*Sec[c + d*x]]) - (7*Tan[c + d*x]^3)/(3*(a + a* Sec[c + d*x])^(3/2)) - (a*Tan[c + d*x]^5)/(a + a*Sec[c + d*x])^(5/2) - (a^ 2*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2))))/d
3.2.67.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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 = 13.54 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.46
| method | result | size |
| default | \(-\frac {a^{2} \left (21 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}+34 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-98 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+70 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+42 \csc \left (d x +c \right )-42 \cot \left (d x +c \right )\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{21 d \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right )^{3} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )^{3}}\) | \(233\) |
-1/21/d*a^2*(21*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^ (1/2)*(-cot(d*x+c)+csc(d*x+c)))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(7/2)+34 *(1-cos(d*x+c))^7*csc(d*x+c)^7-98*(1-cos(d*x+c))^5*csc(d*x+c)^5+70*(1-cos( d*x+c))^3*csc(d*x+c)^3+42*csc(d*x+c)-42*cot(d*x+c))*(-2*a/((1-cos(d*x+c))^ 2*csc(d*x+c)^2-1))^(1/2)/(csc(d*x+c)-cot(d*x+c)+1)^3/(-cot(d*x+c)+csc(d*x+ c)-1)^3
Time = 0.34 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.34 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=\left [\frac {21 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (10 \, a^{2} \cos \left (d x + c\right )^{3} - 16 \, a^{2} \cos \left (d x + c\right )^{2} - 12 \, a^{2} \cos \left (d x + c\right ) - 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (21 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (10 \, a^{2} \cos \left (d x + c\right )^{3} - 16 \, a^{2} \cos \left (d x + c\right )^{2} - 12 \, a^{2} \cos \left (d x + c\right ) - 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{21 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
[1/21*(21*(a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*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*(10*a^2*cos( d*x + c)^3 - 16*a^2*cos(d*x + c)^2 - 12*a^2*cos(d*x + c) - 3*a^2)*sqrt((a* cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d* x + c)^3), 2/21*(21*(a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*sqrt(a)*arct an(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (10*a^2*cos(d*x + c)^3 - 16*a^2*cos(d*x + c)^2 - 12*a^2*cos(d*x + c) - 3*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d *x + c)^4 + d*cos(d*x + c)^3)]
\[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \tan ^{2}{\left (c + d x \right )}\, dx \]
\[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{2} \,d x } \]
1/42*(21*((a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d *x + 2*c) + a^2)*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) - (a^ 2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a ^2)*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^2*d*cos(2*d *x + 2*c)^2 + a^2*d*sin(2*d*x + 2*c)^2 + 2*a^2*d*cos(2*d*x + 2*c) + a^2*d) *integrate((((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2 *d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*s in(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(9/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2 *cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2 *cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(9/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c))))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2...
\[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^2(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]