3.171 \(\int \frac{\tan ^5(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 (a \sec (c+d x)+a)^{7/2}}{7 a^4 d}-\frac{6 (a \sec (c+d x)+a)^{5/2}}{5 a^3 d}+\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a^2 d}+\frac{2 \sqrt{a \sec (c+d x)+a}}{a d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.101379, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 (a \sec (c+d x)+a)^{7/2}}{7 a^4 d}-\frac{6 (a \sec (c+d x)+a)^{5/2}}{5 a^3 d}+\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a^2 d}+\frac{2 \sqrt{a \sec (c+d x)+a}}{a d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) + (2*Sqrt[a + a*Sec[c + d*x]])/(a*d) + (2*(a + a*Se
c[c + d*x])^(3/2))/(3*a^2*d) - (6*(a + a*Sec[c + d*x])^(5/2))/(5*a^3*d) + (2*(a + a*Sec[c + d*x])^(7/2))/(7*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 \frac{\tan ^5(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(-a+a x)^2 (a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-3 a^2 (a+a x)^{3/2}+\frac{a^2 (a+a x)^{3/2}}{x}+a (a+a x)^{5/2}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=-\frac{6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac{2 (a+a \sec (c+d x))^{7/2}}{7 a^4 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d}-\frac{6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac{2 (a+a \sec (c+d x))^{7/2}}{7 a^4 d}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac{2 \sqrt{a+a \sec (c+d x)}}{a d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d}-\frac{6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac{2 (a+a \sec (c+d x))^{7/2}}{7 a^4 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{2 \sqrt{a+a \sec (c+d x)}}{a d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d}-\frac{6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac{2 (a+a \sec (c+d x))^{7/2}}{7 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 d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}+\frac{2 \sqrt{a+a \sec (c+d x)}}{a d}+\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d}-\frac{6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac{2 (a+a \sec (c+d x))^{7/2}}{7 a^4 d}\\ \end{align*}

Mathematica [A]  time = 0.18377, size = 88, normalized size = 0.7 \[ \frac{2 \left (15 \sec ^4(c+d x)-3 \sec ^3(c+d x)-64 \sec ^2(c+d x)+46 \sec (c+d x)-105 \sqrt{\sec (c+d x)+1} \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )+92\right )}{105 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*(92 + 46*Sec[c + d*x] - 64*Sec[c + d*x]^2 - 3*Sec[c + d*x]^3 + 15*Sec[c + d*x]^4 - 105*ArcTanh[Sqrt[1 + Sec
[c + d*x]]]*Sqrt[1 + Sec[c + d*x]]))/(105*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.239, size = 293, normalized size = 2.3 \begin{align*} -{\frac{1}{840\,ad \left ( \cos \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 105\, \left ( \cos \left ( dx+c \right ) \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 ) ^{7/2}+315\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\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 ) ^{7/2}+315\,\cos \left ( dx+c \right ) \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 ) ^{7/2}+105\,\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 ) ^{7/2}-1472\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+736\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+288\,\cos \left ( dx+c \right ) -240 \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))^(1/2),x)

[Out]

-1/840/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(105*cos(d*x+c)^3*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)+315*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))^(7/2)+315*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))^(7/2)+105*2^(1/2)*arctan(1/2*2^(1/2)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)-1472*cos(d*x+c)^3+736*cos(d*x+c)^2+288*co
s(d*x+c)-240)/cos(d*x+c)^3

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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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95474, size = 759, normalized size = 6.02 \begin{align*} \left [\frac{105 \, \sqrt{a} \cos \left (d x + c\right )^{3} \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 (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{210 \, a d \cos \left (d x + c\right )^{3}}, \frac{105 \, \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 )^{3} + 2 \,{\left (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{105 \, a d \cos \left (d x + c\right )^{3}}\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))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(105*sqrt(a)*cos(d*x + c)^3*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*(92*cos(d*x + c)^3 - 46*cos(d*x + c)^2 - 18*cos
(d*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a*d*cos(d*x + c)^3), 1/105*(105*sqrt(-a)*arctan(2*sq
rt(-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)^3 + 2*(92*cos
(d*x + c)^3 - 46*cos(d*x + c)^2 - 18*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a*d*cos(d*x
+ c)^3)]

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

[Out]

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

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Giac [A]  time = 12.2559, size = 257, normalized size = 2.04 \begin{align*} -\frac{\sqrt{2}{\left (\frac{105 \, \sqrt{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} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{2 \,{\left (105 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} - 70 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} a - 252 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} a^{2} - 120 \, a^{3}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )}}{105 \, 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))^(1/2),x, algorithm="giac")

[Out]

-1/105*sqrt(2)*(105*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*sgn(tan
(1/2*d*x + 1/2*c)^2 - 1)) + 2*(105*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3 - 70*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a -
252*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2 - 120*a^3)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d