3.828 \(\int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=126 \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{a d}+\frac {3 \tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

[Out]

arctanh(cos(d*x+c))/a/d-cot(d*x+c)/a/d-sec(d*x+c)/a/d-1/3*sec(d*x+c)^3/a/d-1/5*sec(d*x+c)^5/a/d+3*tan(d*x+c)/a
/d+tan(d*x+c)^3/a/d+1/5*tan(d*x+c)^5/a/d

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2839, 2620, 270, 2622, 302, 207} \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{a d}+\frac {3 \tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

ArcTanh[Cos[c + d*x]]/(a*d) - Cot[c + d*x]/(a*d) - Sec[c + d*x]/(a*d) - Sec[c + d*x]^3/(3*a*d) - Sec[c + d*x]^
5/(5*a*d) + (3*Tan[c + d*x])/(a*d) + Tan[c + d*x]^3/(a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 207

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps

\begin {align*} \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \csc (c+d x) \sec ^6(c+d x) \, dx}{a}+\frac {\int \csc ^2(c+d x) \sec ^6(c+d x) \, dx}{a}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^6}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}+\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^2} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \left (3+\frac {1}{x^2}+3 x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a d}-\frac {\operatorname {Subst}\left (\int \left (1+x^2+x^4+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {\cot (c+d x)}{a d}-\frac {\sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {3 \tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{a d}+\frac {\tan ^5(c+d x)}{5 a d}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {\cot (c+d x)}{a d}-\frac {\sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {3 \tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{a d}+\frac {\tan ^5(c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.60, size = 341, normalized size = 2.71 \[ -\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (352 \sin (c+d x)-596 \sin (2 (c+d x))+864 \sin (3 (c+d x))-298 \sin (4 (c+d x))+384 \sin (5 (c+d x))+1216 \cos (2 (c+d x))+149 \cos (3 (c+d x))+528 \cos (4 (c+d x))+149 \cos (5 (c+d x))+480 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+120 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-120 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-120 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (240 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-240 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-298\right )-480 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-240 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+176\right )}{3840 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

-1/3840*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(176 + 1216*Cos[2*(c + d*x)] + 149*Cos[3*(c + d*x)]
+ 528*Cos[4*(c + d*x)] + 149*Cos[5*(c + d*x)] + 120*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 120*Cos[5*(c + d*
x)]*Log[Cos[(c + d*x)/2]] - 120*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 120*Cos[5*(c + d*x)]*Log[Sin[(c + d*x
)/2]] + Cos[c + d*x]*(-298 - 240*Log[Cos[(c + d*x)/2]] + 240*Log[Sin[(c + d*x)/2]]) + 352*Sin[c + d*x] - 596*S
in[2*(c + d*x)] - 480*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] + 480*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 86
4*Sin[3*(c + d*x)] - 298*Sin[4*(c + d*x)] - 240*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 240*Log[Sin[(c + d*x)
/2]]*Sin[4*(c + d*x)] + 384*Sin[5*(c + d*x)]))/(a*d*(1 + Sin[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 194, normalized size = 1.54 \[ \frac {66 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (48 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 8}{30 \, {\left (a d \cos \left (d x + c\right )^{5} - a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - a d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(66*cos(d*x + c)^4 - 28*cos(d*x + c)^2 + 15*(cos(d*x + c)^5 - cos(d*x + c)^3*sin(d*x + c) - cos(d*x + c)^
3)*log(1/2*cos(d*x + c) + 1/2) - 15*(cos(d*x + c)^5 - cos(d*x + c)^3*sin(d*x + c) - cos(d*x + c)^3)*log(-1/2*c
os(d*x + c) + 1/2) + 2*(48*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - 1)*sin(d*x + c) - 8)/(a*d*cos(d*x + c)^5 - a*d*
cos(d*x + c)^3*sin(d*x + c) - a*d*cos(d*x + c)^3)

________________________________________________________________________________________

giac [A]  time = 0.22, size = 178, normalized size = 1.41 \[ -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {60 \, {\left (2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {5 \, {\left (27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {585 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2040 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2890 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1880 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 493}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - 60*tan(1/2*d*x + 1/2*c)/a - 60*(2*tan(1/2*d*x + 1/2*c) - 1)/(a*
tan(1/2*d*x + 1/2*c)) + 5*(27*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x + 1/2*c) + 25)/(a*(tan(1/2*d*x + 1/2*c)
- 1)^3) + (585*tan(1/2*d*x + 1/2*c)^4 + 2040*tan(1/2*d*x + 1/2*c)^3 + 2890*tan(1/2*d*x + 1/2*c)^2 + 1880*tan(1
/2*d*x + 1/2*c) + 493)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

________________________________________________________________________________________

maple [A]  time = 0.50, size = 223, normalized size = 1.77 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {1}{6 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {9}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2}{5 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {7}{3 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {39}{8 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/2/a/d*tan(1/2*d*x+1/2*c)-1/6/a/d/(tan(1/2*d*x+1/2*c)-1)^3-1/4/a/d/(tan(1/2*d*x+1/2*c)-1)^2-9/8/a/d/(tan(1/2*
d*x+1/2*c)-1)-1/2/a/d/tan(1/2*d*x+1/2*c)-1/a/d*ln(tan(1/2*d*x+1/2*c))-2/5/a/d/(tan(1/2*d*x+1/2*c)+1)^5+1/a/d/(
tan(1/2*d*x+1/2*c)+1)^4-7/3/a/d/(tan(1/2*d*x+1/2*c)+1)^3+5/2/a/d/(tan(1/2*d*x+1/2*c)+1)^2-39/8/a/d/(tan(1/2*d*
x+1/2*c)+1)

________________________________________________________________________________________

maxima [B]  time = 0.34, size = 379, normalized size = 3.01 \[ -\frac {\frac {\frac {122 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {26 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {454 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {252 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {510 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {330 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {210 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {195 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 15}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {6 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {2 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {15 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*((122*sin(d*x + c)/(cos(d*x + c) + 1) - 26*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 454*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 252*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 510*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 330*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6 - 210*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 195*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 + 15)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*a*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 - 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 2*
a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 2*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - a*sin(d*x + c)^9/(cos(d*x +
c) + 1)^9) + 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - 15*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

________________________________________________________________________________________

mupad [B]  time = 10.82, size = 257, normalized size = 2.04 \[ \frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {454\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}+\frac {26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {122\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-1}{d\,\left (-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*sin(c + d*x)^2*(a + a*sin(c + d*x))),x)

[Out]

((26*tan(c/2 + (d*x)/2)^2)/15 - (122*tan(c/2 + (d*x)/2))/15 + (454*tan(c/2 + (d*x)/2)^3)/15 + (84*tan(c/2 + (d
*x)/2)^4)/5 - 34*tan(c/2 + (d*x)/2)^5 - 22*tan(c/2 + (d*x)/2)^6 + 14*tan(c/2 + (d*x)/2)^7 + 13*tan(c/2 + (d*x)
/2)^8 - 1)/(d*(2*a*tan(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 - 4*a*tan(c/2 + (d*x)/2)^3 - 12*a*tan(c/2 + (
d*x)/2)^4 + 12*a*tan(c/2 + (d*x)/2)^6 + 4*a*tan(c/2 + (d*x)/2)^7 - 4*a*tan(c/2 + (d*x)/2)^8 - 2*a*tan(c/2 + (d
*x)/2)^9)) - log(tan(c/2 + (d*x)/2))/(a*d) + tan(c/2 + (d*x)/2)/(2*a*d)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________