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

Optimal. Leaf size=105 \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{\tan ^7(c+d x)}{a^3 d}-\frac{3 \tan ^5(c+d x)}{5 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{\sec ^7(c+d x)}{a^3 d}+\frac{3 \sec ^5(c+d x)}{5 a^3 d} \]

[Out]

(3*Sec[c + d*x]^5)/(5*a^3*d) - Sec[c + d*x]^7/(a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x]^5)/(5*a
^3*d) - Tan[c + d*x]^7/(a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

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Rubi [A]  time = 0.33538, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2875, 2873, 2606, 14, 2607, 270} \[ -\frac{4 \tan ^9(c+d x)}{9 a^3 d}-\frac{\tan ^7(c+d x)}{a^3 d}-\frac{3 \tan ^5(c+d x)}{5 a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{\sec ^7(c+d x)}{a^3 d}+\frac{3 \sec ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[c + d*x]^5)/(5*a^3*d) - Sec[c + d*x]^7/(a^3*d) + (4*Sec[c + d*x]^9)/(9*a^3*d) - (3*Tan[c + d*x]^5)/(5*a
^3*d) - Tan[c + d*x]^7/(a^3*d) - (4*Tan[c + d*x]^9)/(9*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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{\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \sec ^7(c+d x) (a-a \sin (c+d x))^3 \tan ^3(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (a^3 \sec ^7(c+d x) \tan ^3(c+d x)-3 a^3 \sec ^6(c+d x) \tan ^4(c+d x)+3 a^3 \sec ^5(c+d x) \tan ^5(c+d x)-a^3 \sec ^4(c+d x) \tan ^6(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \sec ^7(c+d x) \tan ^3(c+d x) \, dx}{a^3}-\frac{\int \sec ^4(c+d x) \tan ^6(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^6(c+d x) \tan ^4(c+d x) \, dx}{a^3}+\frac{3 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int x^6 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac{3 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}\\ &=\frac{3 \sec ^5(c+d x)}{5 a^3 d}-\frac{\sec ^7(c+d x)}{a^3 d}+\frac{4 \sec ^9(c+d x)}{9 a^3 d}-\frac{3 \tan ^5(c+d x)}{5 a^3 d}-\frac{\tan ^7(c+d x)}{a^3 d}-\frac{4 \tan ^9(c+d x)}{9 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.315734, size = 185, normalized size = 1.76 \[ \frac{4608 \sin (c+d x)-1323 \sin (2 (c+d x))-128 \sin (3 (c+d x))-588 \sin (4 (c+d x))+384 \sin (5 (c+d x))+49 \sin (6 (c+d x))-1764 \cos (c+d x)-4032 \cos (2 (c+d x))-98 \cos (3 (c+d x))+768 \cos (4 (c+d x))+294 \cos (5 (c+d x))-64 \cos (6 (c+d x))+5376}{46080 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5376 - 1764*Cos[c + d*x] - 4032*Cos[2*(c + d*x)] - 98*Cos[3*(c + d*x)] + 768*Cos[4*(c + d*x)] + 294*Cos[5*(c
+ d*x)] - 64*Cos[6*(c + d*x)] + 4608*Sin[c + d*x] - 1323*Sin[2*(c + d*x)] - 128*Sin[3*(c + d*x)] - 588*Sin[4*(
c + d*x)] + 384*Sin[5*(c + d*x)] + 49*Sin[6*(c + d*x)])/(46080*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a + a*Sin[c + d*x])^3)

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Maple [A]  time = 0.131, size = 190, normalized size = 1.8 \begin{align*} 16\,{\frac{1}{d{a}^{3}} \left ( -{\frac{1}{384\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{3}}}-{\frac{1}{256\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}-{\frac{1}{512\,\tan \left ( 1/2\,dx+c/2 \right ) -512}}+1/18\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-9}-1/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}+1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}-{\frac{7}{12\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}+{\frac{67}{160\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}-{\frac{11}{64\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{5}{192\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{1}{128\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{1}{512\,\tan \left ( 1/2\,dx+c/2 \right ) +512}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

16/d/a^3*(-1/384/(tan(1/2*d*x+1/2*c)-1)^3-1/256/(tan(1/2*d*x+1/2*c)-1)^2-1/512/(tan(1/2*d*x+1/2*c)-1)+1/18/(ta
n(1/2*d*x+1/2*c)+1)^9-1/4/(tan(1/2*d*x+1/2*c)+1)^8+1/2/(tan(1/2*d*x+1/2*c)+1)^7-7/12/(tan(1/2*d*x+1/2*c)+1)^6+
67/160/(tan(1/2*d*x+1/2*c)+1)^5-11/64/(tan(1/2*d*x+1/2*c)+1)^4+5/192/(tan(1/2*d*x+1/2*c)+1)^3+1/128/(tan(1/2*d
*x+1/2*c)+1)^2+1/512/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.09962, size = 570, normalized size = 5.43 \begin{align*} \frac{4 \,{\left (\frac{6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{18 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{84 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{54 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{45 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}}{45 \,{\left (a^{3} + \frac{6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac{12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac{6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac{a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

4/45*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 18*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 18*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 84*sin(d*x + c)
^6/(cos(d*x + c) + 1)^6 + 54*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 45*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)
/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x +
c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 + 36*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*
x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11 - a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)

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Fricas [A]  time = 1.68134, size = 323, normalized size = 3.08 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{6} - 9 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} -{\left (6 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 10}{45 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} +{\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\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)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/45*(2*cos(d*x + c)^6 - 9*cos(d*x + c)^4 + 15*cos(d*x + c)^2 - (6*cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 5)*sin(
d*x + c) - 10)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)
^3)*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)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.27307, size = 217, normalized size = 2.07 \begin{align*} -\frac{\frac{15 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 540 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3120 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 5940 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 8298 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6372 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3528 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 972 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 113}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{9}}}{1440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/1440*(15*(3*tan(1/2*d*x + 1/2*c)^2 + 1)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (45*tan(1/2*d*x + 1/2*c)^8 + 5
40*tan(1/2*d*x + 1/2*c)^7 + 3120*tan(1/2*d*x + 1/2*c)^6 + 5940*tan(1/2*d*x + 1/2*c)^5 + 8298*tan(1/2*d*x + 1/2
*c)^4 + 6372*tan(1/2*d*x + 1/2*c)^3 + 3528*tan(1/2*d*x + 1/2*c)^2 + 972*tan(1/2*d*x + 1/2*c) + 113)/(a^3*(tan(
1/2*d*x + 1/2*c) + 1)^9))/d