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

Optimal. Leaf size=149 \[ -\frac {2 \cos (c+d x)}{a^2 d}+\frac {9 \tan ^5(c+d x)}{10 a^2 d}-\frac {3 \tan ^3(c+d x)}{2 a^2 d}+\frac {9 \tan (c+d x)}{2 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}-\frac {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}-\frac {9 x}{2 a^2} \]

[Out]

-9/2*x/a^2-2*cos(d*x+c)/a^2/d-6*sec(d*x+c)/a^2/d+2*sec(d*x+c)^3/a^2/d-2/5*sec(d*x+c)^5/a^2/d+9/2*tan(d*x+c)/a^
2/d-3/2*tan(d*x+c)^3/a^2/d+9/10*tan(d*x+c)^5/a^2/d-1/2*sin(d*x+c)^2*tan(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.31, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2875, 2710, 3473, 8, 2590, 270, 2591, 288, 302, 203} \[ -\frac {2 \cos (c+d x)}{a^2 d}+\frac {9 \tan ^5(c+d x)}{10 a^2 d}-\frac {3 \tan ^3(c+d x)}{2 a^2 d}+\frac {9 \tan (c+d x)}{2 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}-\frac {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}-\frac {9 x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*x)/(2*a^2) - (2*Cos[c + d*x])/(a^2*d) - (6*Sec[c + d*x])/(a^2*d) + (2*Sec[c + d*x]^3)/(a^2*d) - (2*Sec[c +
d*x]^5)/(5*a^2*d) + (9*Tan[c + d*x])/(2*a^2*d) - (3*Tan[c + d*x]^3)/(2*a^2*d) + (9*Tan[c + d*x]^5)/(10*a^2*d)
- (Sin[c + d*x]^2*Tan[c + d*x]^5)/(2*a^2*d)

Rule 8

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 2590

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

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2710

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, 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 ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int (a-a \sin (c+d x))^2 \tan ^6(c+d x) \, dx}{a^4}\\ &=\frac {\int \left (a^2 \tan ^6(c+d x)-2 a^2 \sin (c+d x) \tan ^6(c+d x)+a^2 \sin ^2(c+d x) \tan ^6(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \tan ^6(c+d x) \, dx}{a^2}+\frac {\int \sin ^2(c+d x) \tan ^6(c+d x) \, dx}{a^2}-\frac {2 \int \sin (c+d x) \tan ^6(c+d x) \, dx}{a^2}\\ &=\frac {\tan ^5(c+d x)}{5 a^2 d}-\frac {\int \tan ^4(c+d x) \, dx}{a^2}+\frac {\operatorname {Subst}\left (\int \frac {x^8}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {2 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {\tan ^5(c+d x)}{5 a^2 d}-\frac {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}+\frac {\int \tan ^2(c+d x) \, dx}{a^2}+\frac {2 \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^6}-\frac {3}{x^4}+\frac {3}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}+\frac {7 \operatorname {Subst}\left (\int \frac {x^6}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac {2 \cos (c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 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 {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}-\frac {\int 1 \, dx}{a^2}+\frac {7 \operatorname {Subst}\left (\int \left (1-x^2+x^4-\frac {1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac {x}{a^2}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {9 \tan (c+d x)}{2 a^2 d}-\frac {3 \tan ^3(c+d x)}{2 a^2 d}+\frac {9 \tan ^5(c+d x)}{10 a^2 d}-\frac {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac {9 x}{2 a^2}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {9 \tan (c+d x)}{2 a^2 d}-\frac {3 \tan ^3(c+d x)}{2 a^2 d}+\frac {9 \tan ^5(c+d x)}{10 a^2 d}-\frac {\sin ^2(c+d x) \tan ^5(c+d x)}{2 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 191, normalized size = 1.28 \[ -\frac {250 \sin (c+d x)+720 c \sin (2 (c+d x))+720 d x \sin (2 (c+d x))-824 \sin (2 (c+d x))+351 \sin (3 (c+d x))+5 \sin (5 (c+d x))+10 (90 c+90 d x-103) \cos (c+d x)+544 \cos (2 (c+d x))-180 c \cos (3 (c+d x))-180 d x \cos (3 (c+d x))+206 \cos (3 (c+d x))-20 \cos (4 (c+d x))+500}{160 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/160*(500 + 10*(-103 + 90*c + 90*d*x)*Cos[c + d*x] + 544*Cos[2*(c + d*x)] + 206*Cos[3*(c + d*x)] - 180*c*Cos
[3*(c + d*x)] - 180*d*x*Cos[3*(c + d*x)] - 20*Cos[4*(c + d*x)] + 250*Sin[c + d*x] - 824*Sin[2*(c + d*x)] + 720
*c*Sin[2*(c + d*x)] + 720*d*x*Sin[2*(c + d*x)] + 351*Sin[3*(c + d*x)] + 5*Sin[5*(c + d*x)])/(a^2*d*(Cos[(c + d
*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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fricas [A]  time = 0.46, size = 132, normalized size = 0.89 \[ -\frac {45 \, d x \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{4} - 90 \, d x \cos \left (d x + c\right ) - 78 \, \cos \left (d x + c\right )^{2} - {\left (5 \, \cos \left (d x + c\right )^{4} + 90 \, d x \cos \left (d x + c\right ) + 84 \, \cos \left (d x + c\right )^{2} - 6\right )} \sin \left (d x + c\right ) + 4}{10 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(45*d*x*cos(d*x + c)^3 + 10*cos(d*x + c)^4 - 90*d*x*cos(d*x + c) - 78*cos(d*x + c)^2 - (5*cos(d*x + c)^4
+ 90*d*x*cos(d*x + c) + 84*cos(d*x + c)^2 - 6)*sin(d*x + c) + 4)/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c)
*sin(d*x + c) - 2*a^2*d*cos(d*x + c))

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giac [A]  time = 0.23, size = 160, normalized size = 1.07 \[ -\frac {\frac {90 \, {\left (d x + c\right )}}{a^{2}} + \frac {20 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {5}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 690 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 750 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 181}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(90*(d*x + c)/a^2 + 20*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 4)/((
tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + 5/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) + (155*tan(1/2*d*x + 1/2*c)^4 + 690*ta
n(1/2*d*x + 1/2*c)^3 + 1120*tan(1/2*d*x + 1/2*c)^2 + 750*tan(1/2*d*x + 1/2*c) + 181)/(a^2*(tan(1/2*d*x + 1/2*c
) + 1)^5))/d

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maple [A]  time = 0.47, size = 267, normalized size = 1.79 \[ -\frac {1}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4}{a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {4}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^2/d/(tan(1/2*d*x+1/2*c)-1)-1/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/a^2/d/(1+tan(1/2*d
*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2+1/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/a^2/d/(1+tan(1/2*d
*x+1/2*c)^2)^2-9/d/a^2*arctan(tan(1/2*d*x+1/2*c))-4/5/a^2/d/(tan(1/2*d*x+1/2*c)+1)^5+2/a^2/d/(tan(1/2*d*x+1/2*
c)+1)^4+1/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3-7/2/a^2/d/(tan(1/2*d*x+1/2*c)+1)^2-31/4/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 = 2.83 \[ -\frac {\frac {\frac {211 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {268 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {212 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {84 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {174 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {300 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {300 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {180 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {45 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 64}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*((211*sin(d*x + c)/(cos(d*x + c) + 1) + 268*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 212*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 84*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 174*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 300*sin(
d*x + c)^6/(cos(d*x + c) + 1)^6 - 300*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 180*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 - 45*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 64)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 7*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 + 6*a^2*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 - 6*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 - 7*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) + 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 17.67, size = 172, normalized size = 1.15 \[ \frac {-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {174\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {212\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {268\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {211\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {64}{5}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {9\,x}{2\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((211*tan(c/2 + (d*x)/2))/5 + (268*tan(c/2 + (d*x)/2)^2)/5 + (212*tan(c/2 + (d*x)/2)^3)/5 + (84*tan(c/2 + (d*x
)/2)^4)/5 - (174*tan(c/2 + (d*x)/2)^5)/5 - 60*tan(c/2 + (d*x)/2)^6 - 60*tan(c/2 + (d*x)/2)^7 - 36*tan(c/2 + (d
*x)/2)^8 - 9*tan(c/2 + (d*x)/2)^9 + 64/5)/(a^2*d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^5*(tan(c/2
+ (d*x)/2)^2 + 1)^2) - (9*x)/(2*a^2)

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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)**2*sin(d*x+c)**6/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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