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]

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

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Rubi [A]  time = 0.10, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 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 \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*}

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Mathematica [A]  time = 0.18, size = 88, normalized size = 0.70 \[ \frac {2 \left (15 \sec ^4(c+d x)-3 \sec ^3(c+d x)-64 \sec ^2(c+d x)+46 \sec (c+d x)-105 \sqrt {\sec (c+d x)+1} \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )+92\right )}{105 d \sqrt {a (\sec (c+d x)+1)}} \]

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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fricas [A]  time = 0.79, size = 285, normalized size = 2.26 \[ \left [\frac {105 \, \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{210 \, a d \cos \left (d x + c\right )^{3}}, \frac {105 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{105 \, a d \cos \left (d x + c\right )^{3}}\right ] \]

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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giac [A]  time = 3.77, size = 190, normalized size = 1.51 \[ -\frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {2 \, {\left (105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} - 70 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a - 252 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} - 120 \, a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{105 \, d} \]

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

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maple [B]  time = 1.33, size = 293, normalized size = 2.33 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (105 \left (\cos ^{3}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+315 \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+315 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+105 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-1472 \left (\cos ^{3}\left (d x +c \right )\right )+736 \left (\cos ^{2}\left (d x +c \right )\right )+288 \cos \left (d x +c \right )-240\right )}{840 d \cos \left (d x +c \right )^{3} a} \]

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

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maxima [A]  time = 0.54, size = 129, normalized size = 1.02 \[ \frac {\frac {105 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {30 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{4}} - \frac {126 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{3}} + \frac {70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{2}} + \frac {210 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a}}{105 \, d} \]

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]

1/105*(105*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a)))/sqrt(a) + 30*(a + a/
cos(d*x + c))^(7/2)/a^4 - 126*(a + a/cos(d*x + c))^(5/2)/a^3 + 70*(a + a/cos(d*x + c))^(3/2)/a^2 + 210*sqrt(a
+ a/cos(d*x + c))/a)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

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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