3.272 \(\int \frac{1}{(a \sec (x)+b \tan (x))^5} \, dx\)
Optimal. Leaf size=101 \[ -\frac{\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac{4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac{3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac{4 a}{b^5 (a+b \sin (x))}+\frac{\log (a+b \sin (x))}{b^5} \]
[Out]
Log[a + b*Sin[x]]/b^5 - (a^2 - b^2)^2/(4*b^5*(a + b*Sin[x])^4) + (4*a*(a^2 - b^2))/(3*b^5*(a + b*Sin[x])^3) -
(3*a^2 - b^2)/(b^5*(a + b*Sin[x])^2) + (4*a)/(b^5*(a + b*Sin[x]))
________________________________________________________________________________________
Rubi [A] time = 0.115666, antiderivative size = 101, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used =
{4391, 2668, 697} \[ -\frac{\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac{4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac{3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac{4 a}{b^5 (a+b \sin (x))}+\frac{\log (a+b \sin (x))}{b^5} \]
Antiderivative was successfully verified.
[In]
Int[(a*Sec[x] + b*Tan[x])^(-5),x]
[Out]
Log[a + b*Sin[x]]/b^5 - (a^2 - b^2)^2/(4*b^5*(a + b*Sin[x])^4) + (4*a*(a^2 - b^2))/(3*b^5*(a + b*Sin[x])^3) -
(3*a^2 - b^2)/(b^5*(a + b*Sin[x])^2) + (4*a)/(b^5*(a + b*Sin[x]))
Rule 4391
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]
Rule 2668
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 697
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]
Rubi steps
\begin{align*} \int \frac{1}{(a \sec (x)+b \tan (x))^5} \, dx &=\int \frac{\cos ^5(x)}{(a+b \sin (x))^5} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{(a+x)^5} \, dx,x,b \sin (x)\right )}{b^5}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a^2-b^2\right )^2}{(a+x)^5}-\frac{4 \left (a^3-a b^2\right )}{(a+x)^4}+\frac{2 \left (3 a^2-b^2\right )}{(a+x)^3}-\frac{4 a}{(a+x)^2}+\frac{1}{a+x}\right ) \, dx,x,b \sin (x)\right )}{b^5}\\ &=\frac{\log (a+b \sin (x))}{b^5}-\frac{\left (a^2-b^2\right )^2}{4 b^5 (a+b \sin (x))^4}+\frac{4 a \left (a^2-b^2\right )}{3 b^5 (a+b \sin (x))^3}-\frac{3 a^2-b^2}{b^5 (a+b \sin (x))^2}+\frac{4 a}{b^5 (a+b \sin (x))}\\ \end{align*}
Mathematica [A] time = 0.332491, size = 86, normalized size = 0.85 \[ \frac{\frac{12 b^2 \left (9 a^2+b^2\right ) \sin ^2(x)+8 a b \left (11 a^2+b^2\right ) \sin (x)+2 a^2 b^2+25 a^4+48 a b^3 \sin ^3(x)-3 b^4}{12 (a+b \sin (x))^4}+\log (a+b \sin (x))}{b^5} \]
Antiderivative was successfully verified.
[In]
Integrate[(a*Sec[x] + b*Tan[x])^(-5),x]
[Out]
(Log[a + b*Sin[x]] + (25*a^4 + 2*a^2*b^2 - 3*b^4 + 8*a*b*(11*a^2 + b^2)*Sin[x] + 12*b^2*(9*a^2 + b^2)*Sin[x]^2
+ 48*a*b^3*Sin[x]^3)/(12*(a + b*Sin[x])^4))/b^5
________________________________________________________________________________________
Maple [A] time = 0.113, size = 130, normalized size = 1.3 \begin{align*}{\frac{4\,{a}^{3}}{3\,{b}^{5} \left ( a+b\sin \left ( x \right ) \right ) ^{3}}}-{\frac{4\,a}{3\,{b}^{3} \left ( a+b\sin \left ( x \right ) \right ) ^{3}}}+{\frac{\ln \left ( a+b\sin \left ( x \right ) \right ) }{{b}^{5}}}-3\,{\frac{{a}^{2}}{{b}^{5} \left ( a+b\sin \left ( x \right ) \right ) ^{2}}}+{\frac{1}{{b}^{3} \left ( a+b\sin \left ( x \right ) \right ) ^{2}}}-{\frac{{a}^{4}}{4\,{b}^{5} \left ( a+b\sin \left ( x \right ) \right ) ^{4}}}+{\frac{{a}^{2}}{2\,{b}^{3} \left ( a+b\sin \left ( x \right ) \right ) ^{4}}}-{\frac{1}{4\,b \left ( a+b\sin \left ( x \right ) \right ) ^{4}}}+4\,{\frac{a}{{b}^{5} \left ( a+b\sin \left ( x \right ) \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a*sec(x)+b*tan(x))^5,x)
[Out]
4/3*a^3/b^5/(a+b*sin(x))^3-4/3*a/b^3/(a+b*sin(x))^3+ln(a+b*sin(x))/b^5-3/b^5/(a+b*sin(x))^2*a^2+1/b^3/(a+b*sin
(x))^2-1/4/b^5/(a+b*sin(x))^4*a^4+1/2/b^3/(a+b*sin(x))^4*a^2-1/4/b/(a+b*sin(x))^4+4*a/b^5/(a+b*sin(x))
________________________________________________________________________________________
Maxima [B] time = 1.6462, size = 652, normalized size = 6.46 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="maxima")
[Out]
-2/3*(3*(a^7 - a^3*b^4)*sin(x)/(cos(x) + 1) + 3*(7*a^6*b - 3*a^2*b^5)*sin(x)^2/(cos(x) + 1)^2 + (9*a^7 + 52*a^
5*b^2 - a^3*b^4 - 12*a*b^6)*sin(x)^3/(cos(x) + 1)^3 + 2*(21*a^6*b + 25*a^4*b^3 - 7*a^2*b^5 - 3*b^7)*sin(x)^4/(
cos(x) + 1)^4 + (9*a^7 + 52*a^5*b^2 - a^3*b^4 - 12*a*b^6)*sin(x)^5/(cos(x) + 1)^5 + 3*(7*a^6*b - 3*a^2*b^5)*si
n(x)^6/(cos(x) + 1)^6 + 3*(a^7 - a^3*b^4)*sin(x)^7/(cos(x) + 1)^7)/(a^8*b^4 + 8*a^7*b^5*sin(x)/(cos(x) + 1) +
8*a^7*b^5*sin(x)^7/(cos(x) + 1)^7 + a^8*b^4*sin(x)^8/(cos(x) + 1)^8 + 4*(a^8*b^4 + 6*a^6*b^6)*sin(x)^2/(cos(x)
+ 1)^2 + 8*(3*a^7*b^5 + 4*a^5*b^7)*sin(x)^3/(cos(x) + 1)^3 + 2*(3*a^8*b^4 + 24*a^6*b^6 + 8*a^4*b^8)*sin(x)^4/
(cos(x) + 1)^4 + 8*(3*a^7*b^5 + 4*a^5*b^7)*sin(x)^5/(cos(x) + 1)^5 + 4*(a^8*b^4 + 6*a^6*b^6)*sin(x)^6/(cos(x)
+ 1)^6) + log(a + 2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/b^5 - log(sin(x)^2/(cos(x) + 1)^2 + 1)/
b^5
________________________________________________________________________________________
Fricas [B] time = 2.56074, size = 509, normalized size = 5.04 \begin{align*} \frac{25 \, a^{4} + 110 \, a^{2} b^{2} + 9 \, b^{4} - 12 \,{\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 12 \,{\left (b^{4} \cos \left (x\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} - 4 \,{\left (a b^{3} \cos \left (x\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (x\right )\right )} \log \left (b \sin \left (x\right ) + a\right ) - 8 \,{\left (6 \, a b^{3} \cos \left (x\right )^{2} - 11 \, a^{3} b - 7 \, a b^{3}\right )} \sin \left (x\right )}{12 \,{\left (b^{9} \cos \left (x\right )^{4} + a^{4} b^{5} + 6 \, a^{2} b^{7} + b^{9} - 2 \,{\left (3 \, a^{2} b^{7} + b^{9}\right )} \cos \left (x\right )^{2} - 4 \,{\left (a b^{8} \cos \left (x\right )^{2} - a^{3} b^{6} - a b^{8}\right )} \sin \left (x\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="fricas")
[Out]
1/12*(25*a^4 + 110*a^2*b^2 + 9*b^4 - 12*(9*a^2*b^2 + b^4)*cos(x)^2 + 12*(b^4*cos(x)^4 + a^4 + 6*a^2*b^2 + b^4
- 2*(3*a^2*b^2 + b^4)*cos(x)^2 - 4*(a*b^3*cos(x)^2 - a^3*b - a*b^3)*sin(x))*log(b*sin(x) + a) - 8*(6*a*b^3*cos
(x)^2 - 11*a^3*b - 7*a*b^3)*sin(x))/(b^9*cos(x)^4 + a^4*b^5 + 6*a^2*b^7 + b^9 - 2*(3*a^2*b^7 + b^9)*cos(x)^2 -
4*(a*b^8*cos(x)^2 - a^3*b^6 - a*b^8)*sin(x))
________________________________________________________________________________________
Sympy [A] time = 33.2854, size = 1727, normalized size = 17.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a*sec(x)+b*tan(x))**5,x)
[Out]
Piecewise((12*a**4*log(a*sec(x)/b + tan(x))*sec(x)**4/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3
+ 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 6*a**4*log(tan(x)**2 +
1)*sec(x)**4/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a
*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 16*a**4*sec(x)**4/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*
sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 48*a**3*b*log
(a*sec(x)/b + tan(x))*tan(x)*sec(x)**3/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*
tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 24*a**3*b*log(tan(x)**2 + 1)*tan(x)*se
c(x)**3/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8
*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 52*a**3*b*tan(x)*sec(x)**3/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan
(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 72*a**2*b
**2*log(a*sec(x)/b + tan(x))*tan(x)**2*sec(x)**2/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*
a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 36*a**2*b**2*log(tan(x)**2 +
1)*tan(x)**2*sec(x)**2/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x
)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 56*a**2*b**2*tan(x)**2*sec(x)**2/(12*a**4*b**5*sec(x)
**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*
tan(x)**4) + 2*a**2*b**2*sec(x)**2/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(
x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 48*a*b**3*log(a*sec(x)/b + tan(x))*tan(x)*
*3*sec(x)/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b*
*8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 24*a*b**3*log(tan(x)**2 + 1)*tan(x)**3*sec(x)/(12*a**4*b**5*sec(x)*
*4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*t
an(x)**4) + 20*a*b**3*tan(x)**3*sec(x)/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*
tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 8*a*b**3*tan(x)*sec(x)/(12*a**4*b**5*s
ec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*
b**9*tan(x)**4) + 12*b**4*log(a*sec(x)/b + tan(x))*tan(x)**4/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec
(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 6*b**4*log(tan(x
)**2 + 1)*tan(x)**4/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2
+ 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) + 6*b**4*tan(x)**2/(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*t
an(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)**3*sec(x) + 12*b**9*tan(x)**4) - 3*b**4/
(12*a**4*b**5*sec(x)**4 + 48*a**3*b**6*tan(x)*sec(x)**3 + 72*a**2*b**7*tan(x)**2*sec(x)**2 + 48*a*b**8*tan(x)*
*3*sec(x) + 12*b**9*tan(x)**4), Ne(b, 0)), ((8*tan(x)**5/(15*sec(x)**5) + 4*tan(x)**3/(3*sec(x)**5) + tan(x)/s
ec(x)**5)/a**5, True))
________________________________________________________________________________________
Giac [A] time = 1.13955, size = 123, normalized size = 1.22 \begin{align*} \frac{\log \left ({\left | b \sin \left (x\right ) + a \right |}\right )}{b^{5}} - \frac{25 \, b^{3} \sin \left (x\right )^{4} + 52 \, a b^{2} \sin \left (x\right )^{3} + 42 \, a^{2} b \sin \left (x\right )^{2} - 12 \, b^{3} \sin \left (x\right )^{2} + 12 \, a^{3} \sin \left (x\right ) - 8 \, a b^{2} \sin \left (x\right ) - 2 \, a^{2} b + 3 \, b^{3}}{12 \,{\left (b \sin \left (x\right ) + a\right )}^{4} b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a*sec(x)+b*tan(x))^5,x, algorithm="giac")
[Out]
log(abs(b*sin(x) + a))/b^5 - 1/12*(25*b^3*sin(x)^4 + 52*a*b^2*sin(x)^3 + 42*a^2*b*sin(x)^2 - 12*b^3*sin(x)^2 +
12*a^3*sin(x) - 8*a*b^2*sin(x) - 2*a^2*b + 3*b^3)/((b*sin(x) + a)^4*b^4)