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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 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)}{(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*}

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Mathematica [A]  time = 0.12, 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])])

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fricas [A]  time = 0.48, size = 241, normalized size = 3.09 \[ \left [\frac {3 \, \sqrt {a} \cos \left (d x + c\right ) \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{6 \, a^{3} d \cos \left (d x + c\right )}, \frac {3 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right ) - 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{3 \, a^{3} d \cos \left (d x + c\right )}\right ] \]

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))]

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giac [B]  time = 4.51, size = 140, normalized size = 1.79 \[ -\frac {2 \, {\left (\frac {3 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\sqrt {2} {\left (9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{3 \, d} \]

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

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maple [B]  time = 1.19, size = 155, normalized size = 1.99 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+3 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+32 \cos \left (d x +c \right )-4\right )}{6 d \cos \left (d x +c \right ) a^{3}} \]

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

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maxima [B]  time = 0.46, size = 163, normalized size = 2.09 \[ \frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{4}} - \frac {18 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a^{3}} + \frac {2 \, {\left (4 \, a + \frac {3 \, a}{\cos \left (d x + c\right )}\right )}}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{2}} - \frac {6}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}} - \frac {2}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}}{3 \, d} \]

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]

1/3*(3*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a)))/a^(5/2) + 2*(a + a/cos(d
*x + c))^(3/2)/a^4 - 18*sqrt(a + a/cos(d*x + c))/a^3 + 2*(4*a + 3*a/cos(d*x + c))/((a + a/cos(d*x + c))^(3/2)*
a^2) - 6/(sqrt(a + a/cos(d*x + c))*a^2) - 2/((a + a/cos(d*x + c))^(3/2)*a))/d

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

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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)

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