3.17 \(\int \frac{\sec ^3(x)}{a+b \cot (x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{b^2 \tanh ^{-1}(\sin (x))}{a^3}+\frac{b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a \cos (x)-b \sin (x)}{\sqrt{a^2+b^2}}\right )}{a^3}-\frac{b \sec (x)}{a^2}+\frac{\tanh ^{-1}(\sin (x))}{2 a}+\frac{\tan (x) \sec (x)}{2 a} \]

[Out]

ArcTanh[Sin[x]]/(2*a) + (b^2*ArcTanh[Sin[x]])/a^3 + (b*Sqrt[a^2 + b^2]*ArcTanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2
+ b^2]])/a^3 - (b*Sec[x])/a^2 + (Sec[x]*Tan[x])/(2*a)

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Rubi [A]  time = 0.200749, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3518, 3110, 3768, 3770, 3104, 3074, 206} \[ \frac{b^2 \tanh ^{-1}(\sin (x))}{a^3}+\frac{b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{a \cos (x)-b \sin (x)}{\sqrt{a^2+b^2}}\right )}{a^3}-\frac{b \sec (x)}{a^2}+\frac{\tanh ^{-1}(\sin (x))}{2 a}+\frac{\tan (x) \sec (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Cot[x]),x]

[Out]

ArcTanh[Sin[x]]/(2*a) + (b^2*ArcTanh[Sin[x]])/a^3 + (b*Sqrt[a^2 + b^2]*ArcTanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2
+ b^2]])/a^3 - (b*Sec[x])/a^2 + (Sec[x]*Tan[x])/(2*a)

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

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 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 0.436512, size = 192, normalized size = 2.43 \[ -\frac{8 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \tan \left (\frac{x}{2}\right )-a}{\sqrt{a^2+b^2}}\right )+\sec ^2(x) \left (\left (a^2+2 b^2\right ) \cos (2 x) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )-2 a^2 \sin (x)+a^2 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-a^2 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )+4 a b \cos (x)+2 b^2 \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-2 b^2 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{4 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + b*Cot[x]),x]

[Out]

-(8*b*Sqrt[a^2 + b^2]*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]] + Sec[x]^2*(4*a*b*Cos[x] + a^2*Log[Cos[x/2] -
Sin[x/2]] + 2*b^2*Log[Cos[x/2] - Sin[x/2]] + (a^2 + 2*b^2)*Cos[2*x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2]
+ Sin[x/2]]) - a^2*Log[Cos[x/2] + Sin[x/2]] - 2*b^2*Log[Cos[x/2] + Sin[x/2]] - 2*a^2*Sin[x]))/(4*a^3)

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Maple [B]  time = 0.051, size = 172, normalized size = 2.2 \begin{align*} -{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{2\,a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{2\,a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }-2\,{\frac{b\sqrt{{a}^{2}+{b}^{2}}}{{a}^{3}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/(tan(1/2*x)+1)^2+1/2/a/(tan(1/2*x)+1)-1/a^2/(tan(1/2*x)+1)*b+1/2/a*ln(tan(1/2*x)+1)+1/a^3*ln(tan(1/2*x)
+1)*b^2+1/2/a/(tan(1/2*x)-1)^2+1/2/a/(tan(1/2*x)-1)+1/a^2/(tan(1/2*x)-1)*b-1/2/a*ln(tan(1/2*x)-1)-1/a^3*ln(tan
(1/2*x)-1)*b^2-2*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(2*b*tan(1/2*x)-2*a)/(a^2+b^2)^(1/2))

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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(sec(x)^3/(a+b*cot(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.65758, size = 433, normalized size = 5.48 \begin{align*} \frac{2 \, \sqrt{a^{2} + b^{2}} b \cos \left (x\right )^{2} \log \left (\frac{2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) +{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) -{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 4 \, a b \cos \left (x\right ) + 2 \, a^{2} \sin \left (x\right )}{4 \, a^{3} \cos \left (x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(a^2 + b^2)*b*cos(x)^2*log((2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 - 2*sqrt(a^2 +
b^2)*(a*cos(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) + (a^2 + 2*b^2)*cos(x)^2*log(
sin(x) + 1) - (a^2 + 2*b^2)*cos(x)^2*log(-sin(x) + 1) - 4*a*b*cos(x) + 2*a^2*sin(x))/(a^3*cos(x)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (x \right )}}{a + b \cot{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+b*cot(x)),x)

[Out]

Integral(sec(x)**3/(a + b*cot(x)), x)

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Giac [B]  time = 1.40407, size = 213, normalized size = 2.7 \begin{align*} \frac{{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{2 \, a^{3}} - \frac{{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{2 \, a^{3}} + \frac{{\left (a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3}} + \frac{a \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac{1}{2} \, x\right )^{2} + a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x) + 1))/a^3 - 1/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x) - 1))/a^3 + (a^2*b + b^3
)*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 +
b^2)*a^3) + (a*tan(1/2*x)^3 + 2*b*tan(1/2*x)^2 + a*tan(1/2*x) - 2*b)/((tan(1/2*x)^2 - 1)^2*a^2)