3.779 \(\int \frac{\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{\cos (c+d x)}{a^2 d}-\frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{2 \tan ^3(c+d x)}{3 a^2 d}-\frac{2 \tan (c+d x)}{a^2 d}+\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{5 \sec ^3(c+d x)}{3 a^2 d}+\frac{4 \sec (c+d x)}{a^2 d}+\frac{2 x}{a^2} \]

[Out]

(2*x)/a^2 + Cos[c + d*x]/(a^2*d) + (4*Sec[c + d*x])/(a^2*d) - (5*Sec[c + d*x]^3)/(3*a^2*d) + (2*Sec[c + d*x]^5
)/(5*a^2*d) - (2*Tan[c + d*x])/(a^2*d) + (2*Tan[c + d*x]^3)/(3*a^2*d) - (2*Tan[c + d*x]^5)/(5*a^2*d)

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Rubi [A]  time = 0.277804, antiderivative size = 120, normalized size of antiderivative = 1., 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, 2606, 194, 3473, 8, 2590, 270} \[ \frac{\cos (c+d x)}{a^2 d}-\frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{2 \tan ^3(c+d x)}{3 a^2 d}-\frac{2 \tan (c+d x)}{a^2 d}+\frac{2 \sec ^5(c+d x)}{5 a^2 d}-\frac{5 \sec ^3(c+d x)}{3 a^2 d}+\frac{4 \sec (c+d x)}{a^2 d}+\frac{2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*x)/a^2 + Cos[c + d*x]/(a^2*d) + (4*Sec[c + d*x])/(a^2*d) - (5*Sec[c + d*x]^3)/(3*a^2*d) + (2*Sec[c + d*x]^5
)/(5*a^2*d) - (2*Tan[c + d*x])/(a^2*d) + (2*Tan[c + d*x]^3)/(3*a^2*d) - (2*Tan[c + d*x]^5)/(5*a^2*d)

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 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 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 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 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]

Rule 8

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.563298, size = 148, normalized size = 1.23 \[ \frac{\sec (c+d x) (400 \sin (c+d x)+480 c \sin (2 (c+d x))+480 d x \sin (2 (c+d x))-796 \sin (2 (c+d x))+304 \sin (3 (c+d x))+(600 c+600 d x-995) \cos (c+d x)+376 \cos (2 (c+d x))-120 c \cos (3 (c+d x))-120 d x \cos (3 (c+d x))+199 \cos (3 (c+d x))-30 \cos (4 (c+d x))+550)}{240 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sec[c + d*x]*(550 + (-995 + 600*c + 600*d*x)*Cos[c + d*x] + 376*Cos[2*(c + d*x)] + 199*Cos[3*(c + d*x)] - 120
*c*Cos[3*(c + d*x)] - 120*d*x*Cos[3*(c + d*x)] - 30*Cos[4*(c + d*x)] + 400*Sin[c + d*x] - 796*Sin[2*(c + d*x)]
+ 480*c*Sin[2*(c + d*x)] + 480*d*x*Sin[2*(c + d*x)] + 304*Sin[3*(c + d*x)]))/(240*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.109, size = 169, normalized size = 1.4 \begin{align*} -{\frac{1}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{1}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+{\frac{4}{5\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-2\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-{\frac{1}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{5}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{17}{4\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x)

[Out]

-1/4/d/a^2/(tan(1/2*d*x+1/2*c)-1)+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)+4/d/a^2*arctan(tan(1/2*d*x+1/2*c))+4/5/d/a^
2/(tan(1/2*d*x+1/2*c)+1)^5-2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^4-1/3/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3+5/2/d/a^2/(tan(
1/2*d*x+1/2*c)+1)^2+17/4/d/a^2/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.57332, size = 452, normalized size = 3.77 \begin{align*} \frac{4 \,{\left (\frac{\frac{97 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{108 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{27 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{40 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{85 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{60 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 28}{a^{2} + \frac{4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{6 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

4/15*((97*sin(d*x + c)/(cos(d*x + c) + 1) + 108*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 27*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 40*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 85*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 60*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 28)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x
+ c) + 1) + 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 4*a^2*sin(
d*x + c)^5/(cos(d*x + c) + 1)^5 - 6*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4*a^2*sin(d*x + c)^7/(cos(d*x +
c) + 1)^7 - a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.1041, size = 325, normalized size = 2.71 \begin{align*} \frac{30 \, d x \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{4} - 60 \, d x \cos \left (d x + c\right ) - 62 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (30 \, d x \cos \left (d x + c\right ) + 38 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right ) - 9}{15 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(30*d*x*cos(d*x + c)^3 + 15*cos(d*x + c)^4 - 60*d*x*cos(d*x + c) - 62*cos(d*x + c)^2 - 2*(30*d*x*cos(d*x
+ c) + 38*cos(d*x + c)^2 + 3)*sin(d*x + c) - 9)/(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.28805, size = 204, normalized size = 1.7 \begin{align*} \frac{\frac{120 \,{\left (d x + c\right )}}{a^{2}} - \frac{15 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \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 ) - 1\right )} a^{2}} + \frac{255 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1170 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1960 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1310 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 313}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(120*(d*x + c)/a^2 - 15*(tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 9)/((tan(1/2*d*x + 1/2*c)^3 -
tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1)*a^2) + (255*tan(1/2*d*x + 1/2*c)^4 + 1170*tan(1/2*d*x + 1/2
*c)^3 + 1960*tan(1/2*d*x + 1/2*c)^2 + 1310*tan(1/2*d*x + 1/2*c) + 313)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d