3.830 \(\int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=140 \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan (c+d x)}{a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {2 \sec (c+d x)}{a^2 d}+\frac {x}{a^2} \]

[Out]

x/a^2+2*sec(d*x+c)/a^2/d-2*sec(d*x+c)^3/a^2/d+6/5*sec(d*x+c)^5/a^2/d-2/7*sec(d*x+c)^7/a^2/d-tan(d*x+c)/a^2/d+1
/3*tan(d*x+c)^3/a^2/d-1/5*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 = 0.30, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 2607, 30, 2606, 194, 3473, 8} \[ \frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan (c+d x)}{a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {2 \sec (c+d x)}{a^2 d}+\frac {x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 194

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

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 2607

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

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \sec ^2(c+d x) (a-a \sin (c+d x))^2 \tan ^6(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sec ^2(c+d x) \tan ^6(c+d x)-2 a^2 \sec (c+d x) \tan ^7(c+d x)+a^2 \tan ^8(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^2(c+d x) \tan ^6(c+d x) \, dx}{a^2}+\frac {\int \tan ^8(c+d x) \, dx}{a^2}-\frac {2 \int \sec (c+d x) \tan ^7(c+d x) \, dx}{a^2}\\ &=\frac {\tan ^7(c+d x)}{7 a^2 d}-\frac {\int \tan ^6(c+d x) \, dx}{a^2}+\frac {\operatorname {Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {2 \operatorname {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {\int \tan ^4(c+d x) \, dx}{a^2}-\frac {2 \operatorname {Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}-\frac {\int \tan ^2(c+d x) \, dx}{a^2}\\ &=\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d}+\frac {\int 1 \, dx}{a^2}\\ &=\frac {x}{a^2}+\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\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 [A]  time = 0.57, size = 257, normalized size = 1.84 \[ \frac {2128 \sin (c+d x)+6720 c \sin (2 (c+d x))+6720 d x \sin (2 (c+d x))-9144 \sin (2 (c+d x))+456 \sin (3 (c+d x))+3360 c \sin (4 (c+d x))+3360 d x \sin (4 (c+d x))-4572 \sin (4 (c+d x))+1528 \sin (5 (c+d x))+42 (280 c+280 d x-381) \cos (c+d x)+5504 \cos (2 (c+d x))+2520 c \cos (3 (c+d x))+2520 d x \cos (3 (c+d x))-3429 \cos (3 (c+d x))+2752 \cos (4 (c+d x))-840 c \cos (5 (c+d x))-840 d x \cos (5 (c+d x))+1143 \cos (5 (c+d x))+4032}{13440 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4032 + 42*(-381 + 280*c + 280*d*x)*Cos[c + d*x] + 5504*Cos[2*(c + d*x)] - 3429*Cos[3*(c + d*x)] + 2520*c*Cos[
3*(c + d*x)] + 2520*d*x*Cos[3*(c + d*x)] + 2752*Cos[4*(c + d*x)] + 1143*Cos[5*(c + d*x)] - 840*c*Cos[5*(c + d*
x)] - 840*d*x*Cos[5*(c + d*x)] + 2128*Sin[c + d*x] - 9144*Sin[2*(c + d*x)] + 6720*c*Sin[2*(c + d*x)] + 6720*d*
x*Sin[2*(c + d*x)] + 456*Sin[3*(c + d*x)] - 4572*Sin[4*(c + d*x)] + 3360*c*Sin[4*(c + d*x)] + 3360*d*x*Sin[4*(
c + d*x)] + 1528*Sin[5*(c + d*x)])/(13440*a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^7)

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fricas [A]  time = 0.48, size = 140, normalized size = 1.00 \[ \frac {105 \, d x \cos \left (d x + c\right )^{5} - 210 \, d x \cos \left (d x + c\right )^{3} - 172 \, \cos \left (d x + c\right )^{4} + 86 \, \cos \left (d x + c\right )^{2} - {\left (210 \, d x \cos \left (d x + c\right )^{3} + 191 \, \cos \left (d x + c\right )^{4} - 129 \, \cos \left (d x + c\right )^{2} + 25\right )} \sin \left (d x + c\right ) - 10}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(105*d*x*cos(d*x + c)^5 - 210*d*x*cos(d*x + c)^3 - 172*cos(d*x + c)^4 + 86*cos(d*x + c)^2 - (210*d*x*cos
(d*x + c)^3 + 191*cos(d*x + c)^4 - 129*cos(d*x + c)^2 + 25)*sin(d*x + c) - 10)/(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 = 0.29, size = 155, normalized size = 1.11 \[ \frac {\frac {840 \, {\left (d x + c\right )}}{a^{2}} + \frac {35 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {1365 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 9345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 26600 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 39410 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30261 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11837 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1886}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/840*(840*(d*x + c)/a^2 + 35*(9*tan(1/2*d*x + 1/2*c)^2 - 21*tan(1/2*d*x + 1/2*c) + 10)/(a^2*(tan(1/2*d*x + 1/
2*c) - 1)^3) + (1365*tan(1/2*d*x + 1/2*c)^6 + 9345*tan(1/2*d*x + 1/2*c)^5 + 26600*tan(1/2*d*x + 1/2*c)^4 + 394
10*tan(1/2*d*x + 1/2*c)^3 + 30261*tan(1/2*d*x + 1/2*c)^2 + 11837*tan(1/2*d*x + 1/2*c) + 1886)/(a^2*(tan(1/2*d*
x + 1/2*c) + 1)^7))/d

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maple [A]  time = 0.56, size = 230, normalized size = 1.64 \[ -\frac {1}{12 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {4}{7 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {8}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{12 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {11}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {13}{8 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^6/(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+3/8/a^2/d/(tan(1/2*d*x+1/2*c)-1)+2/d/a
^2*arctan(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)^6-8/5/a^2/d/(t
an(1/2*d*x+1/2*c)+1)^5-1/a^2/d/(tan(1/2*d*x+1/2*c)+1)^4+5/12/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3+11/8/a^2/d/(tan(1/
2*d*x+1/2*c)+1)^2+13/8/a^2/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.43, size = 421, normalized size = 3.01 \[ \frac {2 \, {\left (\frac {\frac {279 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1048 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {364 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1554 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {980 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {280 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {420 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {105 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 96}{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 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

2/105*((279*sin(d*x + c)/(cos(d*x + c) + 1) - 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1048*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 - 364*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1554*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 980*
sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 280*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 420*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8 - 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 96)/(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*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 16.89, size = 156, normalized size = 1.11 \[ \frac {x}{a^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {148\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {104\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {2096\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {88\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {186\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {64}{35}}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(cos(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

x/a^2 + ((88*tan(c/2 + (d*x)/2)^2)/35 - (186*tan(c/2 + (d*x)/2))/35 + (2096*tan(c/2 + (d*x)/2)^3)/105 + (104*t
an(c/2 + (d*x)/2)^4)/15 - (148*tan(c/2 + (d*x)/2)^5)/5 - (56*tan(c/2 + (d*x)/2)^6)/3 + (16*tan(c/2 + (d*x)/2)^
7)/3 + 8*tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^9 - 64/35)/(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(sec(d*x+c)**4*sin(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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