3.3.68 \(\int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\) [268]

3.3.68.1 Optimal result
3.3.68.2 Mathematica [C] (verified)
3.3.68.3 Rubi [A] (verified)
3.3.68.4 Maple [A] (verified)
3.3.68.5 Fricas [B] (verification not implemented)
3.3.68.6 Sympy [F]
3.3.68.7 Maxima [B] (verification not implemented)
3.3.68.8 Giac [B] (verification not implemented)
3.3.68.9 Mupad [B] (verification not implemented)

3.3.68.1 Optimal result

Integrand size = 26, antiderivative size = 231 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {a b^2}{2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))^2}-\frac {2 a b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac {\left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \cos (c+d x)\right ) \csc ^2(c+d x)}{2 \left (a^2-b^2\right )^3 d}+\frac {(2 a-b) \log (1-\cos (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\cos (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^4 d} \]

output
1/2*a*b^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))^2-2*a*b*(a^2+b^2)/(a^2-b^2)^3/d/( 
b+a*cos(d*x+c))-1/2*(a*(a^2+3*b^2)-b*(3*a^2+b^2)*cos(d*x+c))*csc(d*x+c)^2/ 
(a^2-b^2)^3/d+1/4*(2*a-b)*ln(1-cos(d*x+c))/(a+b)^4/d+1/4*(2*a+b)*ln(1+cos( 
d*x+c))/(a-b)^4/d-a*(a^4+8*a^2*b^2+3*b^4)*ln(b+a*cos(d*x+c))/(a^2-b^2)^4/d
 
3.3.68.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.88 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.04 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {a b^2 (b+a \cos (c+d x)) \tan ^3(c+d x)}{2 (-a+b)^2 (a+b)^2 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {2 a b (-i a+b) (i a+b) (b+a \cos (c+d x))^2 \tan ^3(c+d x)}{(-a+b)^3 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {2 i \left (a^5+8 a^3 b^2+3 a b^4\right ) (c+d x) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{(a-b)^4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a-b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {i (2 a+b) \arctan (\tan (c+d x)) (b+a \cos (c+d x))^3 \tan ^3(c+d x)}{2 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}-\frac {(b+a \cos (c+d x))^3 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a+b) (b+a \cos (c+d x))^3 \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (-a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {\left (-a^5-8 a^3 b^2-3 a b^4\right ) (b+a \cos (c+d x))^3 \log (b+a \cos (c+d x)) \tan ^3(c+d x)}{\left (-a^2+b^2\right )^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(2 a-b) (b+a \cos (c+d x))^3 \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{4 (a+b)^4 d (a \sin (c+d x)+b \tan (c+d x))^3}+\frac {(b+a \cos (c+d x))^3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3(c+d x)}{8 (-a+b)^3 d (a \sin (c+d x)+b \tan (c+d x))^3} \]

input
Integrate[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
 
output
(a*b^2*(b + a*Cos[c + d*x])*Tan[c + d*x]^3)/(2*(-a + b)^2*(a + b)^2*d*(a*S 
in[c + d*x] + b*Tan[c + d*x])^3) + (2*a*b*((-I)*a + b)*(I*a + b)*(b + a*Co 
s[c + d*x])^2*Tan[c + d*x]^3)/((-a + b)^3*(a + b)^3*d*(a*Sin[c + d*x] + b* 
Tan[c + d*x])^3) + ((2*I)*(a^5 + 8*a^3*b^2 + 3*a*b^4)*(c + d*x)*(b + a*Cos 
[c + d*x])^3*Tan[c + d*x]^3)/((a - b)^4*(a + b)^4*d*(a*Sin[c + d*x] + b*Ta 
n[c + d*x])^3) - ((I/2)*(2*a - b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x] 
)^3*Tan[c + d*x]^3)/((a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - (( 
I/2)*(2*a + b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x])^3*Tan[c + d*x]^3) 
/((-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) - ((b + a*Cos[c + d*x] 
)^3*Csc[(c + d*x)/2]^2*Tan[c + d*x]^3)/(8*(a + b)^3*d*(a*Sin[c + d*x] + b* 
Tan[c + d*x])^3) + ((2*a + b)*(b + a*Cos[c + d*x])^3*Log[Cos[(c + d*x)/2]^ 
2]*Tan[c + d*x]^3)/(4*(-a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + 
((-a^5 - 8*a^3*b^2 - 3*a*b^4)*(b + a*Cos[c + d*x])^3*Log[b + a*Cos[c + d*x 
]]*Tan[c + d*x]^3)/((-a^2 + b^2)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) 
+ ((2*a - b)*(b + a*Cos[c + d*x])^3*Log[Sin[(c + d*x)/2]^2]*Tan[c + d*x]^3 
)/(4*(a + b)^4*d*(a*Sin[c + d*x] + b*Tan[c + d*x])^3) + ((b + a*Cos[c + d* 
x])^3*Sec[(c + d*x)/2]^2*Tan[c + d*x]^3)/(8*(-a + b)^3*d*(a*Sin[c + d*x] + 
 b*Tan[c + d*x])^3)
 
3.3.68.3 Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4897, 3042, 3316, 27, 601, 25, 2160, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a \cos (c+d x)+b)^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle -\frac {a^3 \int \frac {\cos ^2(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a \int \frac {a^2 \cos ^2(c+d x)}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )^2}d(a \cos (c+d x))}{d}\)

\(\Big \downarrow \) 601

\(\displaystyle -\frac {a \left (\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}-\frac {\int -\frac {-\frac {b \left (3 a^2+b^2\right ) \cos ^3(c+d x) a^5}{\left (a^2-b^2\right )^3}+\frac {\left (2 a^4-3 b^2 a^2-3 b^4\right ) \cos ^2(c+d x) a^4}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (7 a^2-3 b^2\right ) \cos (c+d x) a^3}{\left (a^2-b^2\right )^3}+\frac {b^4 \left (3 a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a \left (\frac {\int \frac {-\frac {b \left (3 a^2+b^2\right ) \cos ^3(c+d x) a^5}{\left (a^2-b^2\right )^3}+\frac {\left (2 a^4-3 b^2 a^2-3 b^4\right ) \cos ^2(c+d x) a^4}{\left (a^2-b^2\right )^3}+\frac {b^3 \left (7 a^2-3 b^2\right ) \cos (c+d x) a^3}{\left (a^2-b^2\right )^3}+\frac {b^4 \left (3 a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3}}{(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}d(a \cos (c+d x))}{2 a^2}+\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 2160

\(\displaystyle -\frac {a \left (\frac {\int \left (-\frac {4 b \left (a^2+b^2\right ) a^2}{\left (a^2-b^2\right )^3 (b+a \cos (c+d x))^2}+\frac {2 b^2 a^2}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^3}+\frac {(2 a-b) a}{2 (a+b)^4 (a-a \cos (c+d x))}-\frac {(2 a+b) a}{2 (a-b)^4 (\cos (c+d x) a+a)}+\frac {2 \left (a^6+8 b^2 a^4+3 b^4 a^2\right )}{\left (a^2-b^2\right )^4 (b+a \cos (c+d x))}\right )d(a \cos (c+d x))}{2 a^2}+\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a \left (\frac {a^2 \left (a^2+3 b^2\right )-a b \left (3 a^2+b^2\right ) \cos (c+d x)}{2 \left (a^2-b^2\right )^3 \left (a^2-a^2 \cos ^2(c+d x)\right )}+\frac {\frac {4 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {a^2 b^2}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}+\frac {2 a^2 \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^4}-\frac {a (2 a-b) \log (a-a \cos (c+d x))}{2 (a+b)^4}-\frac {a (2 a+b) \log (a \cos (c+d x)+a)}{2 (a-b)^4}}{2 a^2}\right )}{d}\)

input
Int[Sec[c + d*x]/(a*Sin[c + d*x] + b*Tan[c + d*x])^3,x]
 
output
-((a*((a^2*(a^2 + 3*b^2) - a*b*(3*a^2 + b^2)*Cos[c + d*x])/(2*(a^2 - b^2)^ 
3*(a^2 - a^2*Cos[c + d*x]^2)) + (-((a^2*b^2)/((a^2 - b^2)^2*(b + a*Cos[c + 
 d*x])^2)) + (4*a^2*b*(a^2 + b^2))/((a^2 - b^2)^3*(b + a*Cos[c + d*x])) - 
(a*(2*a - b)*Log[a - a*Cos[c + d*x]])/(2*(a + b)^4) - (a*(2*a + b)*Log[a + 
 a*Cos[c + d*x]])/(2*(a - b)^4) + (2*a^2*(a^4 + 8*a^2*b^2 + 3*b^4)*Log[b + 
 a*Cos[c + d*x]])/(a^2 - b^2)^4)/(2*a^2)))/d)
 

3.3.68.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 601
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2160
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

rule 4897
Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 
3.3.68.4 Maple [A] (verified)

Time = 14.86 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.85

method result size
derivativedivides \(\frac {-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} a}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +\cos \left (d x +c \right ) a \right )^{2}}-\frac {2 a b \left (a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +\cos \left (d x +c \right ) a \right )}+\frac {1}{4 \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}-\frac {1}{4 \left (a -b \right )^{3} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (2 a +b \right ) \ln \left (\cos \left (d x +c \right )+1\right )}{4 \left (a -b \right )^{4}}}{d}\) \(196\)
default \(\frac {-\frac {a \left (a^{4}+8 a^{2} b^{2}+3 b^{4}\right ) \ln \left (b +\cos \left (d x +c \right ) a \right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{2} a}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +\cos \left (d x +c \right ) a \right )^{2}}-\frac {2 a b \left (a^{2}+b^{2}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +\cos \left (d x +c \right ) a \right )}+\frac {1}{4 \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\cos \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}-\frac {1}{4 \left (a -b \right )^{3} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (2 a +b \right ) \ln \left (\cos \left (d x +c \right )+1\right )}{4 \left (a -b \right )^{4}}}{d}\) \(196\)
risch \(\text {Expression too large to display}\) \(1304\)

input
int(sec(d*x+c)/(sin(d*x+c)*a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
1/d*(-a*(a^4+8*a^2*b^2+3*b^4)/(a+b)^4/(a-b)^4*ln(b+cos(d*x+c)*a)+1/2*b^2/( 
a+b)^2*a/(a-b)^2/(b+cos(d*x+c)*a)^2-2*a*b*(a^2+b^2)/(a+b)^3/(a-b)^3/(b+cos 
(d*x+c)*a)+1/4/(a+b)^3/(cos(d*x+c)-1)+1/4*(2*a-b)/(a+b)^4*ln(cos(d*x+c)-1) 
-1/4/(a-b)^3/(cos(d*x+c)+1)+1/4*(2*a+b)/(a-b)^4*ln(cos(d*x+c)+1))
 
3.3.68.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1076 vs. \(2 (223) = 446\).

Time = 0.42 (sec) , antiderivative size = 1076, normalized size of antiderivative = 4.66 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")
 
output
1/4*(8*a^5*b^2 + 8*a^3*b^4 - 16*a*b^6 - 2*(7*a^6*b - 2*a^4*b^3 - 5*a^2*b^5 
)*cos(d*x + c)^3 + 2*(a^7 - 7*a^5*b^2 - a^3*b^4 + 7*a*b^6)*cos(d*x + c)^2 
+ 2*(6*a^6*b + a^4*b^3 - 8*a^2*b^5 + b^7)*cos(d*x + c) + 4*(a^5*b^2 + 8*a^ 
3*b^4 + 3*a*b^6 - (a^7 + 8*a^5*b^2 + 3*a^3*b^4)*cos(d*x + c)^4 - 2*(a^6*b 
+ 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^3 + (a^7 + 7*a^5*b^2 - 5*a^3*b^4 - 3 
*a*b^6)*cos(d*x + c)^2 + 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c))*l 
og(a*cos(d*x + c) + b) - (2*a^5*b^2 + 9*a^4*b^3 + 16*a^3*b^4 + 14*a^2*b^5 
+ 6*a*b^6 + b^7 - (2*a^7 + 9*a^6*b + 16*a^5*b^2 + 14*a^4*b^3 + 6*a^3*b^4 + 
 a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^ 
4 + 6*a^2*b^5 + a*b^6)*cos(d*x + c)^3 + (2*a^7 + 9*a^6*b + 14*a^5*b^2 + 5* 
a^4*b^3 - 10*a^3*b^4 - 13*a^2*b^5 - 6*a*b^6 - b^7)*cos(d*x + c)^2 + 2*(2*a 
^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4 + 6*a^2*b^5 + a*b^6)*cos(d*x + 
c))*log(1/2*cos(d*x + c) + 1/2) - (2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14 
*a^2*b^5 + 6*a*b^6 - b^7 - (2*a^7 - 9*a^6*b + 16*a^5*b^2 - 14*a^4*b^3 + 6* 
a^3*b^4 - a^2*b^5)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 
14*a^3*b^4 + 6*a^2*b^5 - a*b^6)*cos(d*x + c)^3 + (2*a^7 - 9*a^6*b + 14*a^5 
*b^2 - 5*a^4*b^3 - 10*a^3*b^4 + 13*a^2*b^5 - 6*a*b^6 + b^7)*cos(d*x + c)^2 
 + 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 14*a^3*b^4 + 6*a^2*b^5 - a*b^6)*c 
os(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 
- 4*a^4*b^6 + a^2*b^8)*d*cos(d*x + c)^4 + 2*(a^9*b - 4*a^7*b^3 + 6*a^5*...
 
3.3.68.6 Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

input
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))**3,x)
 
output
Integral(sec(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x))**3, x)
 
3.3.68.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 602 vs. \(2 (223) = 446\).

Time = 0.26 (sec) , antiderivative size = 602, normalized size of antiderivative = 2.61 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=-\frac {\frac {8 \, {\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (a + b - \frac {{\left (a - b\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {4 \, {\left (2 \, a - b\right )} \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {a^{6} - 2 \, a^{5} b - a^{4} b^{2} + 4 \, a^{3} b^{3} - a^{2} b^{4} - 2 \, a b^{5} + b^{6} - \frac {2 \, {\left (a^{6} - 20 \, a^{5} b - 11 \, a^{4} b^{2} - 24 \, a^{3} b^{3} - 29 \, a^{2} b^{4} + 4 \, a b^{5} - b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (a^{6} - 38 \, a^{5} b + 31 \, a^{4} b^{2} - 52 \, a^{3} b^{3} + 63 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{\frac {{\left (a^{9} + a^{8} b - 4 \, a^{7} b^{2} - 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} + 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} - 4 \, a^{2} b^{7} + a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, {\left (a^{9} - a^{8} b - 4 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} - 4 \, a^{3} b^{6} + 4 \, a^{2} b^{7} + a b^{8} - b^{9}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (a^{9} - 3 \, a^{8} b + 8 \, a^{6} b^{3} - 6 \, a^{5} b^{4} - 6 \, a^{4} b^{5} + 8 \, a^{3} b^{6} - 3 \, a b^{8} + b^{9}\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\sin \left (d x + c\right )^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{8 \, d} \]

input
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")
 
output
-1/8*(8*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(a + b - (a - b)*sin(d*x + c)^2/(co 
s(d*x + c) + 1)^2)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) - 4*(2* 
a - b)*log(sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4 
*a*b^3 + b^4) + (a^6 - 2*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + 
 b^6 - 2*(a^6 - 20*a^5*b - 11*a^4*b^2 - 24*a^3*b^3 - 29*a^2*b^4 + 4*a*b^5 
- b^6)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + (a^6 - 38*a^5*b + 31*a^4*b^2 
- 52*a^3*b^3 + 63*a^2*b^4 - 6*a*b^5 + b^6)*sin(d*x + c)^4/(cos(d*x + c) + 
1)^4)/((a^9 + a^8*b - 4*a^7*b^2 - 4*a^6*b^3 + 6*a^5*b^4 + 6*a^4*b^5 - 4*a^ 
3*b^6 - 4*a^2*b^7 + a*b^8 + b^9)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*( 
a^9 - a^8*b - 4*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 6*a^4*b^5 - 4*a^3*b^6 + 
4*a^2*b^7 + a*b^8 - b^9)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (a^9 - 3*a^ 
8*b + 8*a^6*b^3 - 6*a^5*b^4 - 6*a^4*b^5 + 8*a^3*b^6 - 3*a*b^8 + b^9)*sin(d 
*x + c)^6/(cos(d*x + c) + 1)^6) + sin(d*x + c)^2/((a^3 - 3*a^2*b + 3*a*b^2 
 - b^3)*(cos(d*x + c) + 1)^2))/d
 
3.3.68.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 801 vs. \(2 (223) = 446\).

Time = 0.67 (sec) , antiderivative size = 801, normalized size of antiderivative = 3.47 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)/(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")
 
output
1/8*(2*(2*a - b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 
4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 8*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(a 
bs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1) 
/(cos(d*x + c) + 1)))/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) + (a 
 + b - 4*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b*(cos(d*x + c) - 1)/ 
(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b 
^3 + b^4)*(cos(d*x + c) - 1)) + (cos(d*x + c) - 1)/((a^3 - 3*a^2*b + 3*a*b 
^2 - b^3)*(cos(d*x + c) + 1)) + 4*(3*a^7 - 2*a^6*b + 11*a^5*b^2 + 28*a^4*b 
^3 + 9*a^3*b^4 + 6*a^2*b^5 + 9*a*b^6 + 6*a^7*(cos(d*x + c) - 1)/(cos(d*x + 
 c) + 1) - 8*a^6*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 38*a^5*b^2*(cos 
(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*a^4*b^3*(cos(d*x + c) - 1)/(cos(d*x 
+ c) + 1) - 26*a^3*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a^2*b^5* 
(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a*b^6*(cos(d*x + c) - 1)/(cos(d 
*x + c) + 1) + 3*a^7*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6*a^6*b*( 
cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 27*a^5*b^2*(cos(d*x + c) - 1)^2 
/(cos(d*x + c) + 1)^2 - 48*a^4*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1) 
^2 + 33*a^3*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 18*a^2*b^5*(co 
s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 9*a*b^6*(cos(d*x + c) - 1)^2/(cos 
(d*x + c) + 1)^2)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(a + b 
+ a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d...
 
3.3.68.9 Mupad [B] (verification not implemented)

Time = 23.58 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.15 \[ \int \frac {\sec (c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-a^5+37\,a^4\,b+6\,a^3\,b^2+58\,a^2\,b^3-5\,a\,b^4+b^5\right )}{2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}-\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{2\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-21\,a^4\,b+10\,a^3\,b^2-34\,a^2\,b^3+5\,a\,b^4-b^5\right )}{\left (a-b\right )\,\left (a^2+2\,a\,b+b^2\right )}}{d\,\left (\left (4\,a^5-20\,a^4\,b+40\,a^3\,b^2-40\,a^2\,b^3+20\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (-8\,a^5+24\,a^4\,b-16\,a^3\,b^2-16\,a^2\,b^3+24\,a\,b^4-8\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,a^5-4\,a^4\,b-8\,a^3\,b^2+8\,a^2\,b^3+4\,a\,b^4-4\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a-b\right )}{d\,\left (2\,a^4+8\,a^3\,b+12\,a^2\,b^2+8\,a\,b^3+2\,b^4\right )}-\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (a^5+8\,a^3\,b^2+3\,a\,b^4\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )} \]

input
int(1/(cos(c + d*x)*(a*sin(c + d*x) + b*tan(c + d*x))^3),x)
 
output
((tan(c/2 + (d*x)/2)^4*(37*a^4*b - 5*a*b^4 - a^5 + b^5 + 58*a^2*b^3 + 6*a^ 
3*b^2))/(2*(a + b)*(2*a*b + a^2 + b^2)) - (3*a*b^2 - 3*a^2*b + a^3 - b^3)/ 
(2*(a + b)) + (tan(c/2 + (d*x)/2)^2*(5*a*b^4 - 21*a^4*b + a^5 - b^5 - 34*a 
^2*b^3 + 10*a^3*b^2))/((a - b)*(2*a*b + a^2 + b^2)))/(d*(tan(c/2 + (d*x)/2 
)^2*(4*a*b^4 - 4*a^4*b + 4*a^5 - 4*b^5 + 8*a^2*b^3 - 8*a^3*b^2) - tan(c/2 
+ (d*x)/2)^4*(8*a^5 - 24*a^4*b - 24*a*b^4 + 8*b^5 + 16*a^2*b^3 + 16*a^3*b^ 
2) + tan(c/2 + (d*x)/2)^6*(20*a*b^4 - 20*a^4*b + 4*a^5 - 4*b^5 - 40*a^2*b^ 
3 + 40*a^3*b^2))) - tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^3) + (log(tan(c/2 + 
(d*x)/2))*(2*a - b))/(d*(8*a*b^3 + 8*a^3*b + 2*a^4 + 2*b^4 + 12*a^2*b^2)) 
- (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(3*a*b^4 + 
 a^5 + 8*a^3*b^2))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2))