\(\int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx\) [270]

Optimal result

Integrand size = 11, antiderivative size = 51 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=-\frac {\log (a+b \sin (x))}{b^3}+\frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4476, 2747, 711} \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=\frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))}-\frac {\log (a+b \sin (x))}{b^3} \]

[In]

Int[(a*Sec[x] + b*Tan[x])^(-3),x]

[Out]

-(Log[a + b*Sin[x]]/b^3) + (a^2 - b^2)/(2*b^3*(a + b*Sin[x])^2) - (2*a)/(b^3*(a + b*Sin[x]))

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4476

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^3(x)}{(a+b \sin (x))^3} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {b^2-x^2}{(a+x)^3} \, dx,x,b \sin (x)\right )}{b^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{-a-x}+\frac {-a^2+b^2}{(a+x)^3}+\frac {2 a}{(a+x)^2}\right ) \, dx,x,b \sin (x)\right )}{b^3} \\ & = -\frac {\log (a+b \sin (x))}{b^3}+\frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=-\frac {\log (a+b \sin (x))+\frac {3 a^2+b^2+4 a b \sin (x)}{2 (a+b \sin (x))^2}}{b^3} \]

[In]

Integrate[(a*Sec[x] + b*Tan[x])^(-3),x]

[Out]

-((Log[a + b*Sin[x]] + (3*a^2 + b^2 + 4*a*b*Sin[x])/(2*(a + b*Sin[x])^2))/b^3)

Maple [A] (verified)

Time = 7.00 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\ln \left (a +b \sin \left (x \right )\right )}{b^{3}}-\frac {-a^{2}+b^{2}}{2 b^{3} \left (a +b \sin \left (x \right )\right )^{2}}-\frac {2 a}{b^{3} \left (a +b \sin \left (x \right )\right )}\)	\(50\)
risch	\(\frac {i x}{b^{3}}-\frac {2 i \left (3 i a^{2} {\mathrm e}^{2 i x}+i b^{2} {\mathrm e}^{2 i x}+2 b a \,{\mathrm e}^{3 i x}-2 a b \,{\mathrm e}^{i x}\right )}{\left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )^{2} b^{3}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{b}-1\right )}{b^{3}}\)	\(103\)

[In]

int(1/(a*sec(x)+b*tan(x))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=\frac {4 \, a b \sin \left (x\right ) + 3 \, a^{2} + b^{2} - 2 \, {\left (b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}\right )} \log \left (b \sin \left (x\right ) + a\right )}{2 \, {\left (b^{5} \cos \left (x\right )^{2} - 2 \, a b^{4} \sin \left (x\right ) - a^{2} b^{3} - b^{5}\right )}} \]

[In]

integrate(1/(a*sec(x)+b*tan(x))^3,x, algorithm="fricas")

[Out]

1/2*(4*a*b*sin(x) + 3*a^2 + b^2 - 2*(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)*log(b*sin(x) + a))/(b^5*cos(x)^2
- 2*a*b^4*sin(x) - a^2*b^3 - b^5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (46) = 92\).

Time = 1.06 (sec) , antiderivative size = 503, normalized size of antiderivative = 9.86 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=\begin {cases} - \frac {2 a^{2} \log {\left (\frac {a \sec {\left (x \right )}}{b} + \tan {\left (x \right )} \right )} \sec ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} + \frac {a^{2} \log {\left (\tan ^{2}{\left (x \right )} + 1 \right )} \sec ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} - \frac {a^{2} \sec ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} - \frac {4 a b \log {\left (\frac {a \sec {\left (x \right )}}{b} + \tan {\left (x \right )} \right )} \tan {\left (x \right )} \sec {\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} + \frac {2 a b \log {\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan {\left (x \right )} \sec {\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} - \frac {2 b^{2} \log {\left (\frac {a \sec {\left (x \right )}}{b} + \tan {\left (x \right )} \right )} \tan ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} + \frac {b^{2} \log {\left (\tan ^{2}{\left (x \right )} + 1 \right )} \tan ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} + \frac {b^{2} \tan ^{2}{\left (x \right )}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} - \frac {b^{2}}{2 a^{2} b^{3} \sec ^{2}{\left (x \right )} + 4 a b^{4} \tan {\left (x \right )} \sec {\left (x \right )} + 2 b^{5} \tan ^{2}{\left (x \right )}} & \text {for}\: b \neq 0 \\\frac {\frac {2 \tan ^{3}{\left (x \right )}}{3 \sec ^{3}{\left (x \right )}} + \frac {\tan {\left (x \right )}}{\sec ^{3}{\left (x \right )}}}{a^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a*sec(x)+b*tan(x))**3,x)

[Out]

Piecewise((-2*a**2*log(a*sec(x)/b + tan(x))*sec(x)**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5
*tan(x)**2) + a**2*log(tan(x)**2 + 1)*sec(x)**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x
)**2) - a**2*sec(x)**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) - 4*a*b*log(a*sec(x
)/b + tan(x))*tan(x)*sec(x)/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) + 2*a*b*log(ta
n(x)**2 + 1)*tan(x)*sec(x)/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) - 2*b**2*log(a*
sec(x)/b + tan(x))*tan(x)**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) + b**2*log(ta
n(x)**2 + 1)*tan(x)**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) + b**2*tan(x)**2/(2
*a**2*b**3*sec(x)**2 + 4*a*b**4*tan(x)*sec(x) + 2*b**5*tan(x)**2) - b**2/(2*a**2*b**3*sec(x)**2 + 4*a*b**4*tan
(x)*sec(x) + 2*b**5*tan(x)**2), Ne(b, 0)), ((2*tan(x)**3/(3*sec(x)**3) + tan(x)/sec(x)**3)/a**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (49) = 98\).

Time = 0.31 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.94 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=\frac {2 \, {\left (\frac {{\left (a^{3} + a b^{2}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {{\left (a^{3} + a b^{2}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{a^{4} b^{2} + \frac {4 \, a^{3} b^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {4 \, a^{3} b^{3} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {a^{4} b^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {2 \, {\left (a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}} - \frac {\log \left (a + \frac {2 \, b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{b^{3}} + \frac {\log \left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{b^{3}} \]

[In]

integrate(1/(a*sec(x)+b*tan(x))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=-\frac {\log \left ({\left | b \sin \left (x\right ) + a \right |}\right )}{b^{3}} + \frac {3 \, b \sin \left (x\right )^{2} + 2 \, a \sin \left (x\right ) - b}{2 \, {\left (b \sin \left (x\right ) + a\right )}^{2} b^{2}} \]

[In]

integrate(1/(a*sec(x)+b*tan(x))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 29.63 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.08 \[ \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx=\frac {2\,a^3\,b\,\sin \left (x\right )+3\,a^2\,b^2\,{\sin \left (x\right )}^2+2\,a\,b^3\,\sin \left (x\right )+b^4\,{\sin \left (x\right )}^2}{2\,a^4\,b^3+4\,a^3\,b^4\,\sin \left (x\right )+2\,a^2\,b^5\,{\sin \left (x\right )}^2}-\frac {2\,\mathrm {atanh}\left (\frac {b^2+a\,\sin \left (x\right )\,b}{2\,a^2+\sin \left (x\right )\,a\,b-b^2}\right )}{b^3} \]

[In]

int(1/(b*tan(x) + a/cos(x))^3,x)

[Out]

(b^4*sin(x)^2 + 3*a^2*b^2*sin(x)^2 + 2*a*b^3*sin(x) + 2*a^3*b*sin(x))/(2*a^4*b^3 + 2*a^2*b^5*sin(x)^2 + 4*a^3*
b^4*sin(x)) - (2*atanh((b^2 + a*b*sin(x))/(2*a^2 - b^2 + a*b*sin(x))))/b^3