\(\int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [202]

Optimal result

Integrand size = 23, antiderivative size = 113 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-4*2^(1/2)*ar 
ctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+2*ta 
n(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.79 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.66 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {8 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (-4 \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\sqrt {2} \left (\arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\sec (c+d x)}}}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {\sec (c+d x)} \sin (c+d x)\right )\right )}{d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

(8*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*(-4*ArcSin[Tan[(c + d*x)/2]]*Sqrt 
[(1 + Sec[c + d*x])^(-1)]*Sqrt[1 + Sec[c + d*x]] + Sqrt[2]*(ArcTan[Tan[(c 
+ d*x)/2]/Sqrt[(1 + Sec[c + d*x])^(-1)]]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqr 
t[1 + Sec[c + d*x]] + Sqrt[(1 + Cos[c + d*x])^(-1)]*Sqrt[Sec[c + d*x]]*Sin 
[c + d*x])))/(d*Sqrt[Sec[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 381, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 381

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

Input:

Int[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

(-2*(-((ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/Sqrt[a] - 
(2*Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]] 
)])/Sqrt[a])/a^2) - Tan[c + d*x]/(a^2*Sqrt[a + a*Sec[c + d*x]])))/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 381

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
+ 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) 
Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 
2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q 
}, x] && NeQ[b*c - a*d, 0] && GtQ[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 = 1.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.70

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\left (1+\cos \left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )-4 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{d \,a^{3} \left (1+\cos \left (d x +c \right )\right )}\)	\(192\)

Input:

int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d/a^3*(a*(1+sec(d*x+c)))^(1/2)*(-(1+cos(d*x+c))*2^(1/2)*(-2*cos(d*x+c)/( 
1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^ 
2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2))-4*(1+cos(d*x+c))*(-2*cos( 
d*x+c)/(1+cos(d*x+c)))^(1/2)*ln((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d 
*x+c)+csc(d*x+c))+2*sin(d*x+c))/(1+cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 414, normalized size of antiderivative = 3.66 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{a^{3} d \cos \left (d x + c\right ) + a^{3} d}, -\frac {2 \, {\left (\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} - \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{a^{3} d \cos \left (d x + c\right ) + a^{3} d}\right ] \] Input:

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**4/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Integral(tan(c + d*x)**4/(a*(sec(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^4/(a*sec(d*x + c) + a)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.58 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4)/(sec(c + d*x)**3 + 3 
*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3