\(\int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx\) [166]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-1)]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 224 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {2 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {6 a^5 \tan ^5(c+d x)}{d (a+a \sec (c+d x))^{5/2}}+\frac {34 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {14 a^7 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 a^8 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}} \] Output:

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)+2/3*a^4*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2 
)+6*a^5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+34/7*a^6*tan(d*x+c)^7/d/(a+a 
*sec(d*x+c))^(7/2)+14/9*a^7*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)+2/11*a^8 
*tan(d*x+c)^11/d/(a+a*sec(d*x+c))^(11/2)
 

Mathematica [A] (verified)

Time = 7.03 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.67 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt {a (1+\sec (c+d x))} \left (5544 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {11}{2}}(c+d x)-1386 \sin \left (\frac {1}{2} (c+d x)\right )+1584 \sin \left (\frac {3}{2} (c+d x)\right )-1386 \sin \left (\frac {5}{2} (c+d x)\right )-143 \sin \left (\frac {7}{2} (c+d x)\right )-693 \sin \left (\frac {9}{2} (c+d x)\right )-26 \sin \left (\frac {11}{2} (c+d x)\right )\right )}{5544 d} \] Input:

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

Output:

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*(5544*Sqrt 
[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(11/2) - 1386*Sin[(c + d 
*x)/2] + 1584*Sin[(3*(c + d*x))/2] - 1386*Sin[(5*(c + d*x))/2] - 143*Sin[( 
7*(c + d*x))/2] - 693*Sin[(9*(c + d*x))/2] - 26*Sin[(11*(c + d*x))/2]))/(5 
544*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 205, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4375

\(\displaystyle -\frac {2 a^5 \int \frac {\tan ^4(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^4}{(\sec (c+d x) a+a)^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 364

\(\displaystyle -\frac {2 a^5 \int \left (\frac {a^3 \tan ^{10}(c+d x)}{(\sec (c+d x) a+a)^5}+\frac {7 a^2 \tan ^8(c+d x)}{(\sec (c+d x) a+a)^4}+\frac {17 a \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {15 \tan ^4(c+d x)}{(\sec (c+d x) a+a)^2}+\frac {\tan ^2(c+d x)}{a (\sec (c+d x) a+a)}+\frac {1}{a^2 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}-\frac {1}{a^2}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^5 \left (-\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2}}-\frac {a^3 \tan ^{11}(c+d x)}{11 (a \sec (c+d x)+a)^{11/2}}-\frac {7 a^2 \tan ^9(c+d x)}{9 (a \sec (c+d x)+a)^{9/2}}+\frac {\tan (c+d x)}{a^2 \sqrt {a \sec (c+d x)+a}}-\frac {17 a \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {3 \tan ^5(c+d x)}{(a \sec (c+d x)+a)^{5/2}}-\frac {\tan ^3(c+d x)}{3 a (a \sec (c+d x)+a)^{3/2}}\right )}{d}\)

Input:

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

Output:

(-2*a^5*(-(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)) - (3*Tan[c + d*x]^5)/(a + a*Sec[c + d*x])^(5/2) - 
(17*a*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2)) - (7*a^2*Tan[c + d*x] 
^9)/(9*(a + a*Sec[c + d*x])^(9/2)) - (a^3*Tan[c + d*x]^11)/(11*(a + a*Sec[ 
c + d*x])^(11/2))))/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 = 212.23 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.81

method result size
default \(-\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (52 \cos \left (d x +c \right )^{5}+719 \cos \left (d x +c \right )^{4}+366 \cos \left (d x +c \right )^{3}-157 \cos \left (d x +c \right )^{2}-224 \cos \left (d x +c \right )-63\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}-693 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{693 d \left (1+\cos \left (d x +c \right )\right )}\) \(181\)

Input:

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

Output:

-2/693/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((52*cos(d*x+c)^5+719 
*cos(d*x+c)^4+366*cos(d*x+c)^3-157*cos(d*x+c)^2-224*cos(d*x+c)-63)*tan(d*x 
+c)*sec(d*x+c)^4-693*(1+cos(d*x+c))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arc 
tanh(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 = 0.11 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.90 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=\left [\frac {693 \, {\left (a^{2} \cos \left (d x + c\right )^{6} + a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (52 \, a^{2} \cos \left (d x + c\right )^{5} + 719 \, a^{2} \cos \left (d x + c\right )^{4} + 366 \, a^{2} \cos \left (d x + c\right )^{3} - 157 \, a^{2} \cos \left (d x + c\right )^{2} - 224 \, a^{2} \cos \left (d x + c\right ) - 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}, -\frac {2 \, {\left (693 \, {\left (a^{2} \cos \left (d x + c\right )^{6} + a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (52 \, a^{2} \cos \left (d x + c\right )^{5} + 719 \, a^{2} \cos \left (d x + c\right )^{4} + 366 \, a^{2} \cos \left (d x + c\right )^{3} - 157 \, a^{2} \cos \left (d x + c\right )^{2} - 224 \, a^{2} \cos \left (d x + c\right ) - 63 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{693 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}\right ] \] Input:

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

Output:

[1/693*(693*(a^2*cos(d*x + c)^6 + a^2*cos(d*x + c)^5)*sqrt(-a)*log((2*a*co 
s(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*(52*a^2*co 
s(d*x + c)^5 + 719*a^2*cos(d*x + c)^4 + 366*a^2*cos(d*x + c)^3 - 157*a^2*c 
os(d*x + c)^2 - 224*a^2*cos(d*x + c) - 63*a^2)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5), -2/693*( 
693*(a^2*cos(d*x + c)^6 + a^2*cos(d*x + c)^5)*sqrt(a)*arctan(sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (52*a^2* 
cos(d*x + c)^5 + 719*a^2*cos(d*x + c)^4 + 366*a^2*cos(d*x + c)^3 - 157*a^2 
*cos(d*x + c)^2 - 224*a^2*cos(d*x + c) - 63*a^2)*sqrt((a*cos(d*x + c) + a) 
/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.51 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^4(c+d x) \, dx=-\frac {\frac {693 \, \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (693 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (3927 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (462 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (1782 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (305 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1331 \, \sqrt {2} a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{693 \, d} \] Input:

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

Output:

-1/693*(693*sqrt(-a)*a^3*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*(693*sqrt(2)*a^8*sgn(cos(d*x 
+ c)) - (3927*sqrt(2)*a^8*sgn(cos(d*x + c)) - (462*sqrt(2)*a^8*sgn(cos(d*x 
 + c)) + (1782*sqrt(2)*a^8*sgn(cos(d*x + c)) + (305*sqrt(2)*a^8*sgn(cos(d* 
x + c))*tan(1/2*d*x + 1/2*c)^2 - 1331*sqrt(2)*a^8*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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sq 
rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 

Mupad [F(-1)]

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

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

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

Output:

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