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

Optimal. Leaf size=78 \[ \frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a^4 d}-\frac{6 \sqrt{a \sec (c+d x)+a}}{a^3 d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.090066, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3880, 88, 63, 207} \[ \frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a^4 d}-\frac{6 \sqrt{a \sec (c+d x)+a}}{a^3 d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - (6*Sqrt[a + a*Sec[c + d*x]])/(a^3*d) + (2*(a + a*
Sec[c + d*x])^(3/2))/(3*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 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 \frac{\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(-a+a x)^2}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{3 a^2}{\sqrt{a+a x}}+\frac{a^2}{x \sqrt{a+a x}}+a \sqrt{a+a x}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=-\frac{6 \sqrt{a+a \sec (c+d x)}}{a^3 d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^4 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac{6 \sqrt{a+a \sec (c+d x)}}{a^3 d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^4 d}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{a^3 d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{6 \sqrt{a+a \sec (c+d x)}}{a^3 d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^4 d}\\ \end{align*}

Mathematica [A]  time = 0.110246, size = 69, normalized size = 0.88 \[ \frac{2 \left (\sec ^2(c+d x)-7 \sec (c+d x)-3 \sqrt{\sec (c+d x)+1} \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )-8\right )}{3 a^2 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-8 - 7*Sec[c + d*x] + Sec[c + d*x]^2 - 3*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*Sqrt[1 + Sec[c + d*x]]))/(3*a^2*d
*Sqrt[a*(1 + Sec[c + d*x])])

________________________________________________________________________________________

Maple [B]  time = 0.199, size = 155, normalized size = 2. \begin{align*} -{\frac{1}{6\,d{a}^{3}\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 3\,\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) +3\,\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}+32\,\cos \left ( dx+c \right ) -4 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(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))^(3/2)*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))^(3/2)+32*cos(d*x+c)-4)/cos(d*x+c)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.90927, size = 629, normalized size = 8.06 \begin{align*} \left [\frac{3 \, \sqrt{a} \cos \left (d x + c\right ) \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 \, \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{6 \, a^{3} d \cos \left (d x + c\right )}, \frac{3 \, \sqrt{-a} \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 ) - 2 \, \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{3 \, a^{3} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(a)*cos(d*x + c)*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos
(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) - 4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(8*cos(d*x +
c) - 1))/(a^3*d*cos(d*x + c)), 1/3*(3*sqrt(-a)*arctan(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) - 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(8*cos(d*x + c) - 1))
/(a^3*d*cos(d*x + c))]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 11.3755, size = 189, normalized size = 2.42 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \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} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 7 \, a\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c
)^2 - 1)) - sqrt(2)*(9*a*tan(1/2*d*x + 1/2*c)^2 - 7*a)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d