\(\int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [203]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}+\frac {\tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)} \left (2+\frac {\tan ^2(c+d x)}{1+\sec (c+d x)}\right )} \] Output:

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+3/2*2^(1/2) 
*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+t 
an(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)/(2+tan(d*x+c)^2/(1+sec(d*x+c)))
 

Mathematica [A] (warning: unable to verify)

Time = 2.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.43 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {4 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (3 \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\sqrt {2} \left (-2 \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\sec (c+d x)}}}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\frac {\sqrt {\frac {1}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sec (c+d x)}}\right )\right )}{d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(4*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*(3*ArcSin[Tan[(c + d*x)/2]]*Sqrt[ 
(1 + Sec[c + d*x])^(-1)]*Sqrt[1 + Sec[c + d*x]] + Sqrt[2]*(-2*ArcTan[Tan[( 
c + d*x)/2]/Sqrt[(1 + Sec[c + d*x])^(-1)]]*Sqrt[(1 + Sec[c + d*x])^(-1)]*S 
qrt[1 + Sec[c + d*x]] + (Sqrt[(1 + Cos[c + d*x])^(-1)]*Tan[(c + d*x)/2])/S 
qrt[Sec[c + d*x]])))/(d*Sqrt[Sec[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5 
/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 373, 397, 216}

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.

\(\displaystyle \int \frac {\tan ^2(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-2*(-1/2*((-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/Sq 
rt[a] + (3*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]] 
)])/(Sqrt[2]*Sqrt[a]))/a - Tan[c + d*x]/(2*a*Sqrt[a + a*Sec[c + d*x]]*(2 + 
 (a*Tan[c + d*x]^2)/(a + a*Sec[c + d*x])))))/(a*d)
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

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] (warning: unable to verify)

Time = 0.85 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.23

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-3 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{2 d \,a^{3}}\)	\(166\)

Input:

int(tan(d*x+c)^2/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d/a^3*(a*(1+sec(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*( 
2*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(csc(d*x 
+c)-cot(d*x+c)))+(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+ 
c))-3*ln((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d*x+c)+csc(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.64 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {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 ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\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 ) - 4 \, \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 )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \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 ) - 4 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\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 ) - 2 \, \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 )}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \] Input:

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[-1/4*(3*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log((2*sqr 
t(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x 
 + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos 
(d*x + c) + 1)) + 4*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*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)) - 4*sqrt(( 
a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c))/(a^3*d*cos(d* 
x + c)^2 + 2*a^3*d*cos(d*x + c) + a^3*d), -1/2*(3*sqrt(2)*(cos(d*x + c)^2 
+ 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos 
(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 4*(cos(d*x + c)^2 + 2*co 
s(d*x + c) + 1)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos 
(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + 
c))*cos(d*x + c)*sin(d*x + c))/(a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c 
) + a^3*d)]
 

Sympy [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**2/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Integral(tan(c + d*x)**2/(a*(sec(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^2/(a*sec(d*x + c) + a)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.35 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, a^{3} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/(a^3* 
d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

int(tan(d*x+c)^2/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

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