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

Optimal. Leaf size=142 \[ -\frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{\tan (c+d x)}{a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}-\frac{11 \sec ^5(c+d x)}{5 a^3 d}+\frac{10 \sec ^3(c+d x)}{3 a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}-\frac{x}{a^3} \]

[Out]

-(x/a^3) - (3*Sec[c + d*x])/(a^3*d) + (10*Sec[c + d*x]^3)/(3*a^3*d) - (11*Sec[c + d*x]^5)/(5*a^3*d) + (4*Sec[c
+ d*x]^7)/(7*a^3*d) + Tan[c + d*x]/(a^3*d) - Tan[c + d*x]^3/(3*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) - (4*Tan[c +
d*x]^7)/(7*a^3*d)

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Rubi [A]  time = 0.339515, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2875, 2873, 2606, 270, 2607, 30, 194, 3473, 8} \[ -\frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{\tan (c+d x)}{a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}-\frac{11 \sec ^5(c+d x)}{5 a^3 d}+\frac{10 \sec ^3(c+d x)}{3 a^3 d}-\frac{3 \sec (c+d x)}{a^3 d}-\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^3) - (3*Sec[c + d*x])/(a^3*d) + (10*Sec[c + d*x]^3)/(3*a^3*d) - (11*Sec[c + d*x]^5)/(5*a^3*d) + (4*Sec[c
+ d*x]^7)/(7*a^3*d) + Tan[c + d*x]/(a^3*d) - Tan[c + d*x]^3/(3*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) - (4*Tan[c +
d*x]^7)/(7*a^3*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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \sec ^3(c+d x) (a-a \sin (c+d x))^3 \tan ^5(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (a^3 \sec ^3(c+d x) \tan ^5(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^6(c+d x)+3 a^3 \sec (c+d x) \tan ^7(c+d x)-a^3 \tan ^8(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \sec ^3(c+d x) \tan ^5(c+d x) \, dx}{a^3}-\frac{\int \tan ^8(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx}{a^3}+\frac{3 \int \sec (c+d x) \tan ^7(c+d x) \, dx}{a^3}\\ &=-\frac{\tan ^7(c+d x)}{7 a^3 d}+\frac{\int \tan ^6(c+d x) \, dx}{a^3}+\frac{\operatorname{Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{\int \tan ^4(c+d x) \, dx}{a^3}+\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}\\ &=-\frac{3 \sec (c+d x)}{a^3 d}+\frac{10 \sec ^3(c+d x)}{3 a^3 d}-\frac{11 \sec ^5(c+d x)}{5 a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}-\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\int \tan ^2(c+d x) \, dx}{a^3}\\ &=-\frac{3 \sec (c+d x)}{a^3 d}+\frac{10 \sec ^3(c+d x)}{3 a^3 d}-\frac{11 \sec ^5(c+d x)}{5 a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\tan (c+d x)}{a^3 d}-\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 a^3 d}-\frac{\int 1 \, dx}{a^3}\\ &=-\frac{x}{a^3}-\frac{3 \sec (c+d x)}{a^3 d}+\frac{10 \sec ^3(c+d x)}{3 a^3 d}-\frac{11 \sec ^5(c+d x)}{5 a^3 d}+\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\tan (c+d x)}{a^3 d}-\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \tan ^7(c+d x)}{7 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.822684, size = 214, normalized size = 1.51 \[ -\frac{2688 \sin (c+d x)+11760 c \sin (2 (c+d x))+11760 d x \sin (2 (c+d x))-23282 \sin (2 (c+d x))+5568 \sin (3 (c+d x))-840 c \sin (4 (c+d x))-840 d x \sin (4 (c+d x))+1663 \sin (4 (c+d x))+14 (840 c+840 d x-1663) \cos (c+d x)+6272 \cos (2 (c+d x))-5040 c \cos (3 (c+d x))-5040 d x \cos (3 (c+d x))+9978 \cos (3 (c+d x))-1768 \cos (4 (c+d x))+4200}{6720 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4200 + 14*(-1663 + 840*c + 840*d*x)*Cos[c + d*x] + 6272*Cos[2*(c + d*x)] + 9978*Cos[3*(c + d*x)] - 5040*c*Co
s[3*(c + d*x)] - 5040*d*x*Cos[3*(c + d*x)] - 1768*Cos[4*(c + d*x)] + 2688*Sin[c + d*x] - 23282*Sin[2*(c + d*x)
] + 11760*c*Sin[2*(c + d*x)] + 11760*d*x*Sin[2*(c + d*x)] + 5568*Sin[3*(c + d*x)] + 1663*Sin[4*(c + d*x)] - 84
0*c*Sin[4*(c + d*x)] - 840*d*x*Sin[4*(c + d*x)])/(6720*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^7)

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Maple [A]  time = 0.117, size = 187, normalized size = 1.3 \begin{align*} -{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+{\frac{8}{7\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-7}}-4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}+{\frac{18}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}-{\frac{5}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{7}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{15}{8\,d{a}^{3}} \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))^3,x)

[Out]

-1/8/d/a^3/(tan(1/2*d*x+1/2*c)-1)-2/d/a^3*arctan(tan(1/2*d*x+1/2*c))+8/7/d/a^3/(tan(1/2*d*x+1/2*c)+1)^7-4/d/a^
3/(tan(1/2*d*x+1/2*c)+1)^6+18/5/d/a^3/(tan(1/2*d*x+1/2*c)+1)^5+1/d/a^3/(tan(1/2*d*x+1/2*c)+1)^4-5/6/d/a^3/(tan
(1/2*d*x+1/2*c)+1)^3-7/4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2-15/8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.62881, size = 452, normalized size = 3.18 \begin{align*} -\frac{2 \,{\left (\frac{\frac{711 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{1274 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{469 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{1260 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{1435 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{630 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{105 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 136}{a^{3} + \frac{6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac{105 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{105 \, 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))^3,x, algorithm="maxima")

[Out]

-2/105*((711*sin(d*x + c)/(cos(d*x + c) + 1) + 1274*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 469*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 1260*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1435*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 63
0*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 105*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 136)/(a^3 + 6*a^3*sin(d*x +
c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*a^3*sin(d*x +
c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 105*arctan(sin(d*x + c)/(cos(d*x + c) +
1))/a^3)/d

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Fricas [A]  time = 1.15031, size = 408, normalized size = 2.87 \begin{align*} -\frac{315 \, d x \cos \left (d x + c\right )^{3} + 221 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 417 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (35 \, d x \cos \left (d x + c\right )^{3} - 140 \, d x \cos \left (d x + c\right ) - 116 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) + 60}{105 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) +{\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \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))^3,x, algorithm="fricas")

[Out]

-1/105*(315*d*x*cos(d*x + c)^3 + 221*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 417*cos(d*x + c)^2 + 3*(35*d*x*co
s(d*x + c)^3 - 140*d*x*cos(d*x + c) - 116*cos(d*x + c)^2 + 15)*sin(d*x + c) + 60)/(3*a^3*d*cos(d*x + c)^3 - 4*
a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28277, size = 174, normalized size = 1.23 \begin{align*} -\frac{\frac{840 \,{\left (d x + c\right )}}{a^{3}} + \frac{105}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} + \frac{1575 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 10920 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 31675 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 48160 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36981 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 14392 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2281}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, 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))^3,x, algorithm="giac")

[Out]

-1/840*(840*(d*x + c)/a^3 + 105/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) + (1575*tan(1/2*d*x + 1/2*c)^6 + 10920*tan(1/
2*d*x + 1/2*c)^5 + 31675*tan(1/2*d*x + 1/2*c)^4 + 48160*tan(1/2*d*x + 1/2*c)^3 + 36981*tan(1/2*d*x + 1/2*c)^2
+ 14392*tan(1/2*d*x + 1/2*c) + 2281)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d