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

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

[Out]

1/2*arctanh(sin(x))/a+b^2*arctanh(sin(x))/a^3-b*sec(x)/a^2+b*arctanh((a*cos(x)-b*sin(x))/(a^2+b^2)^(1/2))*(a^2
+b^2)^(1/2)/a^3+1/2*sec(x)*tan(x)/a

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Rubi [A]  time = 0.20, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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]

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*}

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Mathematica [B]  time = 0.47, 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]

-1/4*(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]))/a^3

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fricas [B]  time = 0.72, size = 166, normalized size = 2.10 \[ \frac {2 \, \sqrt {a^{2} + b^{2}} b \cos \relax (x)^{2} \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}}\right ) + {\left (a^{2} + 2 \, b^{2}\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) - {\left (a^{2} + 2 \, b^{2}\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) - 4 \, a b \cos \relax (x) + 2 \, a^{2} \sin \relax (x)}{4 \, a^{3} \cos \relax (x)^{2}} \]

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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giac [B]  time = 1.01, size = 158, normalized size = 2.00 \[ \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{2 \, a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{2 \, a^{3}} + \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} + a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{2} a^{2}} \]

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)

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maple [B]  time = 0.35, size = 172, normalized size = 2.18 \[ -\frac {2 b \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 \tan \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}}+\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )-1\right )}+\frac {b}{a^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a}-\frac {\ln \left (\tan \left (\frac {x}{2}\right )-1\right ) b^{2}}{a^{3}}-\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )+1\right )}-\frac {b}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )+1\right ) b^{2}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.65, size = 203, normalized size = 2.57 \[ -\frac {2 \, b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {a \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{a^{2} - \frac {2 \, a^{2} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {a^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} - 1\right )}{2 \, a^{3}} + \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.58, size = 346, normalized size = 4.38 \[ \frac {\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{a}-\frac {2\,b}{a^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+\frac {\mathrm {atanh}\left (\frac {24\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2\,b+24\,b^3+\frac {16\,b^5}{a^2}}+\frac {16\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^4\,b+24\,a^2\,b^3+16\,b^5}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b+\frac {24\,b^3}{a}+\frac {16\,b^5}{a^3}}\right )\,\left (a^2+2\,b^2\right )}{a^3}-\frac {2\,b\,\mathrm {atanh}\left (\frac {16\,b^3\,\sqrt {a^2+b^2}}{32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )+16\,a\,b^3+\frac {16\,b^5}{a}+32\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {32\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{16\,b^3+\frac {16\,b^5}{a^2}+\frac {32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+32\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {16\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2+16\,a^2\,b^3+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4+16\,b^5}\right )\,\sqrt {a^2+b^2}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + b*cot(x))),x)

[Out]

(tan(x/2)/a - (2*b)/a^2 + tan(x/2)^3/a + (2*b*tan(x/2)^2)/a^2)/(tan(x/2)^4 - 2*tan(x/2)^2 + 1) + (atanh((24*b^
3*tan(x/2))/(8*a^2*b + 24*b^3 + (16*b^5)/a^2) + (16*b^5*tan(x/2))/(8*a^4*b + 16*b^5 + 24*a^2*b^3) + (8*a*b*tan
(x/2))/(8*a*b + (24*b^3)/a + (16*b^5)/a^3))*(a^2 + 2*b^2))/a^3 - (2*b*atanh((16*b^3*(a^2 + b^2)^(1/2))/(32*b^4
*tan(x/2) + 16*a*b^3 + (16*b^5)/a + 32*a^2*b^2*tan(x/2)) + (32*b^2*tan(x/2)*(a^2 + b^2)^(1/2))/(16*b^3 + (16*b
^5)/a^2 + (32*b^4*tan(x/2))/a + 32*a*b^2*tan(x/2)) + (16*b^4*tan(x/2)*(a^2 + b^2)^(1/2))/(16*b^5 + 16*a^2*b^3
+ 32*a^3*b^2*tan(x/2) + 32*a*b^4*tan(x/2)))*(a^2 + b^2)^(1/2))/a^3

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

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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