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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.145893, antiderivative size = 193, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.734564, 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))

________________________________________________________________________________________

Maple [B]  time = 0.316, size = 500, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

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^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(45045*cos(d*x+c)^6*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+270270*cos(d*x+c)^5*2^(1/2)*arctan(1/2*2^(1/2
)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+675675*cos(d*x+c)^4*2^(1/2)*arct
an(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+900900*cos(d*x+c)^3
*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+675675
*cos(d*x+c)^2*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(13/2)+270270*cos(d*x+c)*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(13/2)+45045*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(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*co
s(d*x+c)^2+1088640*cos(d*x+c)+443520)/cos(d*x+c)^6

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.24114, size = 1056, normalized size = 5.47 \begin{align*} \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 ] \end{align*}

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)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 4.85917, size = 332, normalized size = 1.72 \begin{align*} \frac{\sqrt{2}{\left (\frac{45045 \, \sqrt{2} a^{2} \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^{2} - 30030 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{5} a^{3} + 36036 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} a^{4} - 51480 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} a^{5} + 80080 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} a^{6} + 393120 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} a^{7} + 221760 \, a^{8}\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 \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{45045 \, d} \end{align*}

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^2*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^2 - 30030*(a*tan(1/2*d*x + 1/2*c)^2 - a)^5*a^3 + 36036*(a*tan(1/2
*d*x + 1/2*c)^2 - a)^4*a^4 - 51480*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*a^5 + 80080*(a*tan(1/2*d*x + 1/2*c)^2 - a)
^2*a^6 + 393120*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^7 + 221760*a^8)/((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*sgn(cos(d*x + c))/d