\(\int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx\) [155]

Optimal result

Integrand size = 23, antiderivative size = 128 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=-\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 \tan ^3(c+d x)}{d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^4 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}} \] 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*a^3*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2) 
+2/5*a^4*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)
 

Mathematica [A] (verified)

Time = 6.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.76 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (-10 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )\right )}{10 d} \] Input:

Integrate[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^2,x]
 

Output:

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(-10*Sqrt[2] 
*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(5/2) + 5*Sin[(3*(c + d*x)) 
/2] + Sin[(5*(c + d*x))/2]))/(10*d)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 118, 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.

\(\displaystyle \int \tan ^2(c+d x) (a \sec (c+d x)+a)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}dx\)

\(\Big \downarrow \) 4375

\(\displaystyle -\frac {2 a^3 \int \frac {\tan ^2(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^2}{(\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 364

\(\displaystyle -\frac {2 a^3 \int \left (\frac {a \tan ^4(c+d x)}{(\sec (c+d x) a+a)^2}+\frac {3 \tan ^2(c+d x)}{\sec (c+d x) a+a}+\frac {1}{a}-\frac {1}{a \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^3 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2}}-\frac {a \tan ^5(c+d x)}{5 (a \sec (c+d x)+a)^{5/2}}-\frac {\tan ^3(c+d x)}{(a \sec (c+d x)+a)^{3/2}}-\frac {\tan (c+d x)}{a \sqrt {a \sec (c+d x)+a}}\right )}{d}\)

Input:

Int[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^2,x]
 

Output:

(-2*a^3*(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]]) - Tan[c + d*x]^3/(a + a*Sec[c + 
 d*x])^(3/2) - (a*Tan[c + d*x]^5)/(5*(a + a*Sec[c + d*x])^(5/2))))/d
 

Defintions of rubi rules used

rule 364

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])
 

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 4375

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]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12

method	result	size
default	\(-\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\sin \left (d x +c \right )-3 \tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )+5 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{5 d \left (1+\cos \left (d x +c \right )\right )}\)	\(144\)

Input:

int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/5/d*a*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*(-sin(d*x+c)-3*tan(d*x+c) 
-sec(d*x+c)*tan(d*x+c)+5*(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))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.51 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\left [\frac {5 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left (5 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^2,x, algorithm="fricas")
 

Output:

[1/5*(5*(a*cos(d*x + c)^3 + a*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*(a*cos(d*x + c)^2 
+ 3*a*cos(d*x + c) + a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + 
c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 2/5*(5*(a*cos(d*x + c)^3 + a*co 
s(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))) + (a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a 
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 
+ d*cos(d*x + c)^2)]
 

Sympy [F]

\[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate((a+a*sec(d*x+c))**(3/2)*tan(d*x+c)**2,x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**(3/2)*tan(c + d*x)**2, x)
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{2} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^2,x, algorithm="maxima")
 

Output:

1/10*(5*((a*cos(2*d*x + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2* 
c) + a)*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)*co 
s(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (a*cos(2*d*x 
 + 2*c)^2 + a*sin(2*d*x + 2*c)^2 + 2*a*cos(2*d*x + 2*c) + a)*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*d*cos(2*d*x + 2*c)^2 + a*d*s 
in(2*d*x + 2*c)^2 + 2*a*d*cos(2*d*x + 2*c) + a*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(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*sin(4*d*x + 4*c)*sin(2*d*x 
+ 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*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(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*ar 
ctan2(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...
 

Giac [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.75 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\frac {\frac {5 \, \sqrt {-a} a^{2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (\sqrt {2} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 5 \, \sqrt {2} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{5 \, d} \] Input:

integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^2,x, algorithm="giac")
 

Output:

1/5*(5*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*(sqrt(2)*a^4*sgn(cos(d*x + c))*tan 
(1/2*d*x + 1/2*c)^4 - 5*sqrt(2)*a^4*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
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int(tan(c + d*x)^2*(a + a/cos(c + d*x))^(3/2),x)
 

Output:

int(tan(c + d*x)^2*(a + a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^{3/2} \tan ^2(c+d x) \, dx=\sqrt {a}\, a \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}d x +\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}d x \right ) \] Input:

int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^2,x)
                                                                                    
                                                                                    
 

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*tan(c + d*x)**2,x) + int(sqrt(sec(c 
+ d*x) + 1)*sec(c + d*x)*tan(c + d*x)**2,x))