Integrand size = 23, antiderivative size = 194 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {2 a^2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {14 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {10 a^5 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {2 a^6 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}} \] Output:
2*a^(3/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2*a^2*tan(d* x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/3*a^3*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2 )+14/5*a^4*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+10/7*a^5*tan(d*x+c)^7/d/( a+a*sec(d*x+c))^(7/2)+2/9*a^6*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)
Time = 6.88 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.63 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt {a (1+\sec (c+d x))} \left (2520 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {9}{2}}(c+d x)+126 \sin \left (\frac {1}{2} (c+d x)\right )-288 \sin \left (\frac {5}{2} (c+d x)\right )-315 \sin \left (\frac {7}{2} (c+d x)\right )-169 \sin \left (\frac {9}{2} (c+d x)\right )\right )}{2520 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^4,x]
Output:
(a*Sec[(c + d*x)/2]*Sec[c + d*x]^4*Sqrt[a*(1 + Sec[c + d*x])]*(2520*Sqrt[2 ]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(9/2) + 126*Sin[(c + d*x)/ 2] - 288*Sin[(5*(c + d*x))/2] - 315*Sin[(7*(c + d*x))/2] - 169*Sin[(9*(c + d*x))/2]))/(2520*d)
Time = 0.29 (sec) , antiderivative size = 178, 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.
| \(\displaystyle \int \tan ^4(c+d x) (a \sec (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a^4 \int \frac {\tan ^4(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^3}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle -\frac {2 a^4 \int \left (\frac {a^2 \tan ^8(c+d x)}{(\sec (c+d x) a+a)^4}+\frac {5 a \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {7 \tan ^4(c+d x)}{(\sec (c+d x) a+a)^2}+\frac {\tan ^2(c+d x)}{a (\sec (c+d x) a+a)}+\frac {1}{a^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}-\frac {1}{a^2}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^4 \left (-\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2}}-\frac {a^2 \tan ^9(c+d x)}{9 (a \sec (c+d x)+a)^{9/2}}+\frac {\tan (c+d x)}{a^2 \sqrt {a \sec (c+d x)+a}}-\frac {5 a \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {7 \tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}-\frac {\tan ^3(c+d x)}{3 a (a \sec (c+d x)+a)^{3/2}}\right )}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^4,x]
Output:
(-2*a^4*(-(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)) - (7*Tan[c + d*x]^5)/(5*(a + a*Sec[c + d*x])^(5/2) ) - (5*a*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2)) - (a^2*Tan[c + d*x ]^9)/(9*(a + a*Sec[c + d*x])^(9/2))))/d
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 = 1.24 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.87
| method | result | size |
| default | \(\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-169 \cos \left (d x +c \right )^{4}-242 \cos \left (d x +c \right )^{3}-24 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+35\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}+315 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{315 d \left (1+\cos \left (d x +c \right )\right )}\) | \(169\) |
Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
2/315/d*a*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((-169*cos(d*x+c)^4-242* cos(d*x+c)^3-24*cos(d*x+c)^2+85*cos(d*x+c)+35)*tan(d*x+c)*sec(d*x+c)^3+315 *(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 = 371, normalized size of antiderivative = 1.91 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\left [\frac {315 \, {\left (a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (169 \, a \cos \left (d x + c\right )^{4} + 242 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 85 \, a \cos \left (d x + c\right ) - 35 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, -\frac {2 \, {\left (315 \, {\left (a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (169 \, a \cos \left (d x + c\right )^{4} + 242 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 85 \, a \cos \left (d x + c\right ) - 35 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^4,x, algorithm="fricas")
Output:
[1/315*(315*(a*cos(d*x + c)^5 + a*cos(d*x + c)^4)*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*(169*a*cos(d*x + c)^4 + 242*a*cos(d*x + c)^3 + 24*a*cos(d*x + c)^2 - 85*a*cos(d*x + c) - 35*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4), -2/315*(315*(a*cos(d*x + c)^5 + a*cos(d*x + c)^4 )*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr t(a)*sin(d*x + c))) + (169*a*cos(d*x + c)^4 + 242*a*cos(d*x + c)^3 + 24*a* cos(d*x + c)^2 - 85*a*cos(d*x + c) - 35*a)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sin(d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)]
\[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*tan(d*x+c)**4,x)
Output:
Integral((a*(sec(c + d*x) + 1))**(3/2)*tan(c + d*x)**4, x)
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^4,x, algorithm="maxima")
Output:
Timed out
Time = 1.01 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.60 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=-\frac {\frac {315 \, \sqrt {-a} a^{2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (315 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1470 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (756 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (\sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 162 \, \sqrt {2} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{315 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^4,x, algorithm="giac")
Output:
-1/315*(315*sqrt(-a)*a^2*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(- a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a) *tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2) *abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + 2*(315*sqrt(2)*a^6*sgn(cos(d*x + c)) - (1470*sqrt(2)*a^6*sgn(cos(d*x + c)) - (756*sqrt(2)*a^6*sgn(cos(d*x + c)) + (sqrt(2)*a^6*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 162*sqrt( 2)*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)* tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int(tan(c + d*x)^4*(a + a/cos(c + d*x))^(3/2),x)
Output:
int(tan(c + d*x)^4*(a + a/cos(c + d*x))^(3/2), x)
\[ \int (a+a \sec (c+d x))^{3/2} \tan ^4(c+d x) \, dx=\sqrt {a}\, a \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}d x +\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{4}d x \right ) \] Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^4,x)
Output:
sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4,x) + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)*tan(c + d*x)**4,x))