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

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

[Out]

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

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Rubi [A]  time = 0.0582467, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3880, 51, 63, 207} \[ \frac{2}{a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{2}{3 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(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) + 2/(3*a*d*(a + a*Sec[c + d*x])^(3/2)) + 2/(a^2*d*S
qrt[a + a*Sec[c + d*x]])

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{2}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac{2}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{2}{a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{2}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{2}{a^2 d \sqrt{a+a \sec (c+d x)}}+\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{2}{3 a d (a+a \sec (c+d x))^{3/2}}+\frac{2}{a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0507304, size = 40, normalized size = 0.51 \[ \frac{2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\sec (c+d x)+1\right )}{3 a d (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + Sec[c + d*x]])/(3*a*d*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [A]  time = 0.035, size = 62, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{{a}^{5/2}}{\it Artanh} \left ({\frac{\sqrt{a+a\sec \left ( dx+c \right ) }}{\sqrt{a}}} \right ) }+2\,{\frac{1}{{a}^{2}\sqrt{a+a\sec \left ( dx+c \right ) }}}+{\frac{2}{3\,a} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92842, size = 826, normalized size = 10.59 \begin{align*} \left [\frac{3 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \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 (4 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac{3 \,{\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \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 ) + 2 \,{\left (4 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 9.57922, size = 180, normalized size = 2.31 \begin{align*} -\frac{\frac{12 \, \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 (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a^{8} + 6 \, \sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{9}}{a^{12} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*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)*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^8 + 6*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a
)*a^9)/(a^12*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d