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

Optimal. Leaf size=193 \[ -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 (a \sec (c+d x)+a)^{13/2}}{13 a^4 d}-\frac {6 (a \sec (c+d x)+a)^{11/2}}{11 a^3 d}+\frac {2 (a \sec (c+d x)+a)^{9/2}}{9 a^2 d}+\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}+\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d}+\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

-2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d+2/3*a*(a+a*sec(d*x+c))^(3/2)/d+2/5*(a+a*sec(d*x+c))^(5/2)
/d+2/7*(a+a*sec(d*x+c))^(7/2)/a/d+2/9*(a+a*sec(d*x+c))^(9/2)/a^2/d-6/11*(a+a*sec(d*x+c))^(11/2)/a^3/d+2/13*(a+
a*sec(d*x+c))^(13/2)/a^4/d+2*a^2*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3880, 88, 50, 63, 207} \[ \frac {2 (a \sec (c+d x)+a)^{13/2}}{13 a^4 d}-\frac {6 (a \sec (c+d x)+a)^{11/2}}{11 a^3 d}+\frac {2 (a \sec (c+d x)+a)^{9/2}}{9 a^2 d}+\frac {2 a^2 \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a d}+\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 d}+\frac {2 a (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d + (2*a^2*Sqrt[a + a*Sec[c + d*x]])/d + (2*a*(a + a*Se
c[c + d*x])^(3/2))/(3*d) + (2*(a + a*Sec[c + d*x])^(5/2))/(5*d) + (2*(a + a*Sec[c + d*x])^(7/2))/(7*a*d) + (2*
(a + a*Sec[c + d*x])^(9/2))/(9*a^2*d) - (6*(a + a*Sec[c + d*x])^(11/2))/(11*a^3*d) + (2*(a + a*Sec[c + d*x])^(
13/2))/(13*a^4*d)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 207

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

Rule 3880

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(d*b^(m - 1)
)^(-1), Subst[Int[((-a + b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x, x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(-a+a x)^2 (a+a x)^{9/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a^2 (a+a x)^{9/2}+\frac {a^2 (a+a x)^{9/2}}{x}+a (a+a x)^{11/2}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{9/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{7/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {a \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}\\ &=-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 156, normalized size = 0.81 \[ \frac {(a (\sec (c+d x)+1))^{5/2} \left (\frac {2}{13} (\sec (c+d x)+1)^{13/2}-\frac {6}{11} (\sec (c+d x)+1)^{11/2}+\frac {2}{9} (\sec (c+d x)+1)^{9/2}+\frac {2}{7} (\sec (c+d x)+1)^{7/2}+\frac {2}{5} (\sec (c+d x)+1)^{5/2}+\frac {2}{3} (\sec (c+d x)+1)^{3/2}+2 \sqrt {\sec (c+d x)+1}-2 \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )\right )}{d (\sec (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sec[c + d*x]))^(5/2)*(-2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + 2*Sqrt[1 + Sec[c + d*x]] + (2*(1 + Sec[c +
d*x])^(3/2))/3 + (2*(1 + Sec[c + d*x])^(5/2))/5 + (2*(1 + Sec[c + d*x])^(7/2))/7 + (2*(1 + Sec[c + d*x])^(9/2
))/9 - (6*(1 + Sec[c + d*x])^(11/2))/11 + (2*(1 + Sec[c + d*x])^(13/2))/13))/(d*(1 + Sec[c + d*x])^(5/2))

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fricas [A]  time = 0.73, size = 386, normalized size = 2.00 \[ \left [\frac {45045 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{6} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (71689 \, a^{2} \cos \left (d x + c\right )^{6} + 31723 \, a^{2} \cos \left (d x + c\right )^{5} - 12531 \, a^{2} \cos \left (d x + c\right )^{4} - 27095 \, a^{2} \cos \left (d x + c\right )^{3} - 4445 \, a^{2} \cos \left (d x + c\right )^{2} + 8505 \, a^{2} \cos \left (d x + c\right ) + 3465 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{90090 \, d \cos \left (d x + c\right )^{6}}, \frac {45045 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{6} + 2 \, {\left (71689 \, a^{2} \cos \left (d x + c\right )^{6} + 31723 \, a^{2} \cos \left (d x + c\right )^{5} - 12531 \, a^{2} \cos \left (d x + c\right )^{4} - 27095 \, a^{2} \cos \left (d x + c\right )^{3} - 4445 \, a^{2} \cos \left (d x + c\right )^{2} + 8505 \, a^{2} \cos \left (d x + c\right ) + 3465 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{45045 \, d \cos \left (d x + c\right )^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/90090*(45045*a^(5/2)*cos(d*x + c)^6*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 4*(71689*a^2*cos(d*x + c)^6 + 31723*a^2*cos(d
*x + c)^5 - 12531*a^2*cos(d*x + c)^4 - 27095*a^2*cos(d*x + c)^3 - 4445*a^2*cos(d*x + c)^2 + 8505*a^2*cos(d*x +
c) + 3465*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^6), 1/45045*(45045*sqrt(-a)*a^2*arcta
n(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a))*cos(d*x + c)^6 + 2*(
71689*a^2*cos(d*x + c)^6 + 31723*a^2*cos(d*x + c)^5 - 12531*a^2*cos(d*x + c)^4 - 27095*a^2*cos(d*x + c)^3 - 44
45*a^2*cos(d*x + c)^2 + 8505*a^2*cos(d*x + c) + 3465*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x
+ c)^6)]

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giac [A]  time = 9.54, size = 244, normalized size = 1.26 \[ \frac {\sqrt {2} {\left (\frac {45045 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} a - 30030 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} a^{2} + 36036 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} a^{3} - 51480 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{4} + 80080 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{5} + 393120 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{6} + 221760 \, a^{7}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{45045 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45045*sqrt(2)*(45045*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2
*(45045*(a*tan(1/2*d*x + 1/2*c)^2 - a)^6*a - 30030*(a*tan(1/2*d*x + 1/2*c)^2 - a)^5*a^2 + 36036*(a*tan(1/2*d*x
+ 1/2*c)^2 - a)^4*a^3 - 51480*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*a^4 + 80080*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a
^5 + 393120*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^6 + 221760*a^7)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^6*sqrt(-a*tan(1/2
*d*x + 1/2*c)^2 + a)))*a^2*sgn(cos(d*x + c))/d

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maple [B]  time = 1.32, size = 500, normalized size = 2.59 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (45045 \left (\cos ^{6}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+270270 \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+675675 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+900900 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+675675 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+270270 \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+45045 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {13}{2}}+9176192 \left (\cos ^{6}\left (d x +c \right )\right )+4060544 \left (\cos ^{5}\left (d x +c \right )\right )-1603968 \left (\cos ^{4}\left (d x +c \right )\right )-3468160 \left (\cos ^{3}\left (d x +c \right )\right )-568960 \left (\cos ^{2}\left (d x +c \right )\right )+1088640 \cos \left (d x +c \right )+443520\right ) a^{2}}{2882880 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2882880/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(45045*cos(d*x+c)^6*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)+270270*cos(d*x+c)^5*2^(1/2)*arctan(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)+675675*cos(d*x+c)^4*2^(1/2)*arctan(1
/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)+900900*cos(d*x+c)^3*2^(
1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)+675675*cos
(d*x+c)^2*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/
2)+270270*cos(d*x+c)*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(13/2)+45045*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(13/2)+9176192*cos(d*x+c)^6+4060544*cos(d*x+c)^5-1603968*cos(d*x+c)^4-3468160*cos(d*x+c)^3-568960*cos(d*
x+c)^2+1088640*cos(d*x+c)+443520)/cos(d*x+c)^6*a^2

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maxima [A]  time = 0.60, size = 181, normalized size = 0.94 \[ \frac {45045 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 18018 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} + \frac {6930 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {13}{2}}}{a^{4}} - \frac {24570 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {11}{2}}}{a^{3}} + \frac {10010 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{2}} + \frac {12870 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a} + 30030 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 90090 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{45045 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45045*(45045*a^(5/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a))) + 18018*
(a + a/cos(d*x + c))^(5/2) + 6930*(a + a/cos(d*x + c))^(13/2)/a^4 - 24570*(a + a/cos(d*x + c))^(11/2)/a^3 + 10
010*(a + a/cos(d*x + c))^(9/2)/a^2 + 12870*(a + a/cos(d*x + c))^(7/2)/a + 30030*(a + a/cos(d*x + c))^(3/2)*a +
90090*sqrt(a + a/cos(d*x + c))*a^2)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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