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

Optimal. Leaf size=149 \[ -\frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {6 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {\sec ^5(c+d x)}{5 a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {\sec (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

[Out]

-arctanh(cos(d*x+c))/a^2/d+sec(d*x+c)/a^2/d+1/3*sec(d*x+c)^3/a^2/d+1/5*sec(d*x+c)^5/a^2/d+2/7*sec(d*x+c)^7/a^2
/d-2*tan(d*x+c)/a^2/d-2*tan(d*x+c)^3/a^2/d-6/5*tan(d*x+c)^5/a^2/d-2/7*tan(d*x+c)^7/a^2/d

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Rubi [A]  time = 0.25, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2875, 2873, 3767, 2622, 302, 207, 2606, 30} \[ -\frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {6 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {\sec ^5(c+d x)}{5 a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {\sec (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Cos[c + d*x]]/(a^2*d)) + Sec[c + d*x]/(a^2*d) + Sec[c + d*x]^3/(3*a^2*d) + Sec[c + d*x]^5/(5*a^2*d)
+ (2*Sec[c + d*x]^7)/(7*a^2*d) - (2*Tan[c + d*x])/(a^2*d) - (2*Tan[c + d*x]^3)/(a^2*d) - (6*Tan[c + d*x]^5)/(5
*a^2*d) - (2*Tan[c + d*x]^7)/(7*a^2*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 2606

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

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 2873

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

Rule 2875

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\csc (c+d x) \sec ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \csc (c+d x) \sec ^8(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (-2 a^2 \sec ^8(c+d x)+a^2 \csc (c+d x) \sec ^8(c+d x)+a^2 \sec ^7(c+d x) \tan (c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \csc (c+d x) \sec ^8(c+d x) \, dx}{a^2}+\frac {\int \sec ^7(c+d x) \tan (c+d x) \, dx}{a^2}-\frac {2 \int \sec ^8(c+d x) \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int x^6 \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {x^8}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac {2 \operatorname {Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=\frac {\sec ^7(c+d x)}{7 a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {6 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {\operatorname {Subst}\left (\int \left (1+x^2+x^4+x^6+\frac {1}{-1+x^2}\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {\sec (c+d x)}{a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {\sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {6 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\sec (c+d x)}{a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}+\frac {\sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {2 \tan (c+d x)}{a^2 d}-\frac {2 \tan ^3(c+d x)}{a^2 d}-\frac {6 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 0.61, size = 352, normalized size = 2.36 \[ \frac {2464 \sin (c+d x)-4472 \sin (2 (c+d x))+2208 \sin (3 (c+d x))-2236 \sin (4 (c+d x))+384 \sin (5 (c+d x))+5312 \cos (2 (c+d x))-1677 \cos (3 (c+d x))+696 \cos (4 (c+d x))+559 \cos (5 (c+d x))+3360 \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+1680 \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1260 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+420 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-14 \cos (c+d x) \left (-420 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+420 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+559\right )+1260 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-420 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3360 \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-1680 \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6216}{6720 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6216 + 5312*Cos[2*(c + d*x)] - 1677*Cos[3*(c + d*x)] + 696*Cos[4*(c + d*x)] + 559*Cos[5*(c + d*x)] - 1260*Cos
[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 420*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 14*Cos[c + d*x]*(559 + 420*
Log[Cos[(c + d*x)/2]] - 420*Log[Sin[(c + d*x)/2]]) + 1260*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 420*Cos[5*(
c + d*x)]*Log[Sin[(c + d*x)/2]] + 2464*Sin[c + d*x] - 4472*Sin[2*(c + d*x)] - 3360*Log[Cos[(c + d*x)/2]]*Sin[2
*(c + d*x)] + 3360*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 2208*Sin[3*(c + d*x)] - 2236*Sin[4*(c + d*x)] - 16
80*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] + 1680*Log[Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] + 384*Sin[5*(c + d*x)]
)/(6720*a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^7)

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fricas [A]  time = 0.46, size = 200, normalized size = 1.34 \[ -\frac {174 \, \cos \left (d x + c\right )^{4} + 158 \, \cos \left (d x + c\right )^{2} + 105 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 105 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, {\left (48 \, \cos \left (d x + c\right )^{4} + 33 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) + 50}{210 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(174*cos(d*x + c)^4 + 158*cos(d*x + c)^2 + 105*(cos(d*x + c)^5 - 2*cos(d*x + c)^3*sin(d*x + c) - 2*cos(
d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) - 105*(cos(d*x + c)^5 - 2*cos(d*x + c)^3*sin(d*x + c) - 2*cos(d*x + c)
^3)*log(-1/2*cos(d*x + c) + 1/2) + 4*(48*cos(d*x + c)^4 + 33*cos(d*x + c)^2 + 5)*sin(d*x + c) + 50)/(a^2*d*cos
(d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3)

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giac [A]  time = 0.22, size = 161, normalized size = 1.08 \[ \frac {\frac {840 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {35 \, {\left (12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {3780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 18585 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 41755 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 51730 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 37506 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14917 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2671}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(840*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 35*(12*tan(1/2*d*x + 1/2*c)^2 - 21*tan(1/2*d*x + 1/2*c) + 11)/
(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) + (3780*tan(1/2*d*x + 1/2*c)^6 + 18585*tan(1/2*d*x + 1/2*c)^5 + 41755*tan(1
/2*d*x + 1/2*c)^4 + 51730*tan(1/2*d*x + 1/2*c)^3 + 37506*tan(1/2*d*x + 1/2*c)^2 + 14917*tan(1/2*d*x + 1/2*c) +
2671)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.76, size = 229, normalized size = 1.54 \[ -\frac {1}{12 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {4}{7 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {22}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {6}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {79}{12 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {39}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {9}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-1/8/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2/a^2/d/(tan(1/2*d*x+1/2*c)-1)+1/d/a
^2*ln(tan(1/2*d*x+1/2*c))+4/7/d/a^2/(tan(1/2*d*x+1/2*c)+1)^7-2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^6+22/5/a^2/d/(tan(
1/2*d*x+1/2*c)+1)^5-6/a^2/d/(tan(1/2*d*x+1/2*c)+1)^4+79/12/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3-39/8/a^2/d/(tan(1/2*
d*x+1/2*c)+1)^2+9/2/a^2/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.34, size = 422, normalized size = 2.83 \[ \frac {\frac {2 \, {\left (\frac {554 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {258 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1108 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1204 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {504 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {1470 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {420 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {315 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {210 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 191\right )}}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(2*(554*sin(d*x + c)/(cos(d*x + c) + 1) + 258*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1108*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 - 1204*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 504*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 14
70*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 420*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 315*sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8 - 210*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 191)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3
*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^2*sin(d*x + c)^4/(
cos(d*x + c) + 1)^4 + 14*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 -
3*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^2*sin(d*x + c)^10/(
cos(d*x + c) + 1)^10) + 105*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 12.34, size = 169, normalized size = 1.13 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-\frac {344\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {2216\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {172\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {1108\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {382}{105}}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a^2*d) - ((1108*tan(c/2 + (d*x)/2))/105 + (172*tan(c/2 + (d*x)/2)^2)/35 - (2216*tan(c
/2 + (d*x)/2)^3)/105 - (344*tan(c/2 + (d*x)/2)^4)/15 + (48*tan(c/2 + (d*x)/2)^5)/5 + 28*tan(c/2 + (d*x)/2)^6 +
8*tan(c/2 + (d*x)/2)^7 - 6*tan(c/2 + (d*x)/2)^8 - 4*tan(c/2 + (d*x)/2)^9 + 382/105)/(a^2*d*(tan(c/2 + (d*x)/2
) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^7)

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

[Out]

Timed out

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