3.7 \(\int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx\)

Optimal. Leaf size=46 \[ \frac {\sec ^4(x)}{4 a}+\frac {\tanh ^{-1}(\sin (x))}{8 a}-\frac {\tan (x) \sec ^3(x)}{4 a}+\frac {\tan (x) \sec (x)}{8 a} \]

[Out]

1/8*arctanh(sin(x))/a+1/4*sec(x)^4/a+1/8*sec(x)*tan(x)/a-1/4*sec(x)^3*tan(x)/a

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3872, 2835, 2606, 30, 2611, 3768, 3770} \[ \frac {\sec ^4(x)}{4 a}+\frac {\tanh ^{-1}(\sin (x))}{8 a}-\frac {\tan (x) \sec ^3(x)}{4 a}+\frac {\tan (x) \sec (x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + a*Csc[x]),x]

[Out]

ArcTanh[Sin[x]]/(8*a) + Sec[x]^4/(4*a) + (Sec[x]*Tan[x])/(8*a) - (Sec[x]^3*Tan[x])/(4*a)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sec ^3(x)}{a+a \csc (x)} \, dx &=\int \frac {\sec ^2(x) \tan (x)}{a+a \sin (x)} \, dx\\ &=\frac {\int \sec ^4(x) \tan (x) \, dx}{a}-\frac {\int \sec ^3(x) \tan ^2(x) \, dx}{a}\\ &=-\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\int \sec ^3(x) \, dx}{4 a}+\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,\sec (x)\right )}{a}\\ &=\frac {\sec ^4(x)}{4 a}+\frac {\sec (x) \tan (x)}{8 a}-\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\int \sec (x) \, dx}{8 a}\\ &=\frac {\tanh ^{-1}(\sin (x))}{8 a}+\frac {\sec ^4(x)}{4 a}+\frac {\sec (x) \tan (x)}{8 a}-\frac {\sec ^3(x) \tan (x)}{4 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 25, normalized size = 0.54 \[ \frac {\frac {1}{1-\sin (x)}+\frac {1}{(\sin (x)+1)^2}+\tanh ^{-1}(\sin (x))}{8 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + a*Csc[x]),x]

[Out]

(ArcTanh[Sin[x]] + (1 - Sin[x])^(-1) + (1 + Sin[x])^(-2))/(8*a)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 70, normalized size = 1.52 \[ -\frac {2 \, \cos \relax (x)^{2} - {\left (\cos \relax (x)^{2} \sin \relax (x) + \cos \relax (x)^{2}\right )} \log \left (\sin \relax (x) + 1\right ) + {\left (\cos \relax (x)^{2} \sin \relax (x) + \cos \relax (x)^{2}\right )} \log \left (-\sin \relax (x) + 1\right ) - 2 \, \sin \relax (x) - 6}{16 \, {\left (a \cos \relax (x)^{2} \sin \relax (x) + a \cos \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+a*csc(x)),x, algorithm="fricas")

[Out]

-1/16*(2*cos(x)^2 - (cos(x)^2*sin(x) + cos(x)^2)*log(sin(x) + 1) + (cos(x)^2*sin(x) + cos(x)^2)*log(-sin(x) +
1) - 2*sin(x) - 6)/(a*cos(x)^2*sin(x) + a*cos(x)^2)

________________________________________________________________________________________

giac [A]  time = 0.57, size = 48, normalized size = 1.04 \[ \frac {\log \left (\sin \relax (x) + 1\right )}{16 \, a} - \frac {\log \left (-\sin \relax (x) + 1\right )}{16 \, a} - \frac {\sin \relax (x)^{2} + \sin \relax (x) + 2}{8 \, a {\left (\sin \relax (x) + 1\right )}^{2} {\left (\sin \relax (x) - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+a*csc(x)),x, algorithm="giac")

[Out]

1/16*log(sin(x) + 1)/a - 1/16*log(-sin(x) + 1)/a - 1/8*(sin(x)^2 + sin(x) + 2)/(a*(sin(x) + 1)^2*(sin(x) - 1))

________________________________________________________________________________________

maple [A]  time = 0.52, size = 44, normalized size = 0.96 \[ -\frac {1}{8 a \left (-1+\sin \relax (x )\right )}-\frac {\ln \left (-1+\sin \relax (x )\right )}{16 a}+\frac {1}{8 a \left (1+\sin \relax (x )\right )^{2}}+\frac {\ln \left (1+\sin \relax (x )\right )}{16 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3/(a+a*csc(x)),x)

[Out]

-1/8/a/(-1+sin(x))-1/16/a*ln(-1+sin(x))+1/8/a/(1+sin(x))^2+1/16*ln(1+sin(x))/a

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 54, normalized size = 1.17 \[ -\frac {\sin \relax (x)^{2} + \sin \relax (x) + 2}{8 \, {\left (a \sin \relax (x)^{3} + a \sin \relax (x)^{2} - a \sin \relax (x) - a\right )}} + \frac {\log \left (\sin \relax (x) + 1\right )}{16 \, a} - \frac {\log \left (\sin \relax (x) - 1\right )}{16 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+a*csc(x)),x, algorithm="maxima")

[Out]

-1/8*(sin(x)^2 + sin(x) + 2)/(a*sin(x)^3 + a*sin(x)^2 - a*sin(x) - a) + 1/16*log(sin(x) + 1)/a - 1/16*log(sin(
x) - 1)/a

________________________________________________________________________________________

mupad [B]  time = 0.22, size = 44, normalized size = 0.96 \[ \frac {\mathrm {atanh}\left (\sin \relax (x)\right )}{8\,a}+\frac {\frac {{\sin \relax (x)}^2}{8}+\frac {\sin \relax (x)}{8}+\frac {1}{4}}{-a\,{\sin \relax (x)}^3-a\,{\sin \relax (x)}^2+a\,\sin \relax (x)+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + a/sin(x))),x)

[Out]

atanh(sin(x))/(8*a) + (sin(x)/8 + sin(x)^2/8 + 1/4)/(a - a*sin(x)^2 - a*sin(x)^3 + a*sin(x))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{3}{\relax (x )}}{\csc {\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+a*csc(x)),x)

[Out]

Integral(sec(x)**3/(csc(x) + 1), x)/a

________________________________________________________________________________________