3.849 \(\int \frac{\sec (c+d x) \tan ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=145 \[ -\frac{8 \tan ^{11}(c+d x)}{11 a^4 d}-\frac{20 \tan ^9(c+d x)}{9 a^4 d}-\frac{16 \tan ^7(c+d x)}{7 a^4 d}-\frac{4 \tan ^5(c+d x)}{5 a^4 d}+\frac{8 \sec ^{11}(c+d x)}{11 a^4 d}-\frac{16 \sec ^9(c+d x)}{9 a^4 d}+\frac{9 \sec ^7(c+d x)}{7 a^4 d}-\frac{\sec ^5(c+d x)}{5 a^4 d} \]

[Out]

-Sec[c + d*x]^5/(5*a^4*d) + (9*Sec[c + d*x]^7)/(7*a^4*d) - (16*Sec[c + d*x]^9)/(9*a^4*d) + (8*Sec[c + d*x]^11)
/(11*a^4*d) - (4*Tan[c + d*x]^5)/(5*a^4*d) - (16*Tan[c + d*x]^7)/(7*a^4*d) - (20*Tan[c + d*x]^9)/(9*a^4*d) - (
8*Tan[c + d*x]^11)/(11*a^4*d)

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Rubi [A]  time = 0.40922, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 18, 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} \[ -\frac{8 \tan ^{11}(c+d x)}{11 a^4 d}-\frac{20 \tan ^9(c+d x)}{9 a^4 d}-\frac{16 \tan ^7(c+d x)}{7 a^4 d}-\frac{4 \tan ^5(c+d x)}{5 a^4 d}+\frac{8 \sec ^{11}(c+d x)}{11 a^4 d}-\frac{16 \sec ^9(c+d x)}{9 a^4 d}+\frac{9 \sec ^7(c+d x)}{7 a^4 d}-\frac{\sec ^5(c+d x)}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*Tan[c + d*x]^3)/(a + a*Sin[c + d*x])^4,x]

[Out]

-Sec[c + d*x]^5/(5*a^4*d) + (9*Sec[c + d*x]^7)/(7*a^4*d) - (16*Sec[c + d*x]^9)/(9*a^4*d) + (8*Sec[c + d*x]^11)
/(11*a^4*d) - (4*Tan[c + d*x]^5)/(5*a^4*d) - (16*Tan[c + d*x]^7)/(7*a^4*d) - (20*Tan[c + d*x]^9)/(9*a^4*d) - (
8*Tan[c + d*x]^11)/(11*a^4*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))^4} \, dx &=\frac{\int \sec ^9(c+d x) (a-a \sin (c+d x))^4 \tan ^3(c+d x) \, dx}{a^8}\\ &=\frac{\int \left (a^4 \sec ^9(c+d x) \tan ^3(c+d x)-4 a^4 \sec ^8(c+d x) \tan ^4(c+d x)+6 a^4 \sec ^7(c+d x) \tan ^5(c+d x)-4 a^4 \sec ^6(c+d x) \tan ^6(c+d x)+a^4 \sec ^5(c+d x) \tan ^7(c+d x)\right ) \, dx}{a^8}\\ &=\frac{\int \sec ^9(c+d x) \tan ^3(c+d x) \, dx}{a^4}+\frac{\int \sec ^5(c+d x) \tan ^7(c+d x) \, dx}{a^4}-\frac{4 \int \sec ^8(c+d x) \tan ^4(c+d x) \, dx}{a^4}-\frac{4 \int \sec ^6(c+d x) \tan ^6(c+d x) \, dx}{a^4}+\frac{6 \int \sec ^7(c+d x) \tan ^5(c+d x) \, dx}{a^4}\\ &=\frac{\operatorname{Subst}\left (\int x^8 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac{\operatorname{Subst}\left (\int x^4 \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac{4 \operatorname{Subst}\left (\int x^6 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac{4 \operatorname{Subst}\left (\int x^4 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac{6 \operatorname{Subst}\left (\int x^6 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-x^4+3 x^6-3 x^8+x^{10}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac{\operatorname{Subst}\left (\int \left (-x^8+x^{10}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac{4 \operatorname{Subst}\left (\int \left (x^6+2 x^8+x^{10}\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac{4 \operatorname{Subst}\left (\int \left (x^4+3 x^6+3 x^8+x^{10}\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac{6 \operatorname{Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=-\frac{\sec ^5(c+d x)}{5 a^4 d}+\frac{9 \sec ^7(c+d x)}{7 a^4 d}-\frac{16 \sec ^9(c+d x)}{9 a^4 d}+\frac{8 \sec ^{11}(c+d x)}{11 a^4 d}-\frac{4 \tan ^5(c+d x)}{5 a^4 d}-\frac{16 \tan ^7(c+d x)}{7 a^4 d}-\frac{20 \tan ^9(c+d x)}{9 a^4 d}-\frac{8 \tan ^{11}(c+d x)}{11 a^4 d}\\ \end{align*}

Mathematica [A]  time = 0.457422, size = 166, normalized size = 1.14 \[ \frac{\sec ^3(c+d x) (844800 \sin (c+d x)-191752 \sin (2 (c+d x))+11264 \sin (3 (c+d x))-69728 \sin (4 (c+d x))+25600 \sin (5 (c+d x))+17432 \sin (6 (c+d x))-1024 \sin (7 (c+d x))-215721 \cos (c+d x)-619520 \cos (2 (c+d x))+23969 \cos (3 (c+d x))+32768 \cos (4 (c+d x))+54475 \cos (5 (c+d x))-8192 \cos (6 (c+d x))-2179 \cos (7 (c+d x))+844800)}{7096320 a^4 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*Tan[c + d*x]^3)/(a + a*Sin[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^3*(844800 - 215721*Cos[c + d*x] - 619520*Cos[2*(c + d*x)] + 23969*Cos[3*(c + d*x)] + 32768*Cos[4
*(c + d*x)] + 54475*Cos[5*(c + d*x)] - 8192*Cos[6*(c + d*x)] - 2179*Cos[7*(c + d*x)] + 844800*Sin[c + d*x] - 1
91752*Sin[2*(c + d*x)] + 11264*Sin[3*(c + d*x)] - 69728*Sin[4*(c + d*x)] + 25600*Sin[5*(c + d*x)] + 17432*Sin[
6*(c + d*x)] - 1024*Sin[7*(c + d*x)]))/(7096320*a^4*d*(1 + Sin[c + d*x])^4)

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Maple [A]  time = 0.131, size = 220, normalized size = 1.5 \begin{align*} 16\,{\frac{1}{d{a}^{4}} \left ( -{\frac{1}{768\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{3}}}-{\frac{1}{512\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}-{\frac{1}{512\,\tan \left ( 1/2\,dx+c/2 \right ) -512}}+1/11\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-11}-1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-10}+{\frac{23}{18\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{9}}}-2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}+{\frac{235}{112\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{7}}}-{\frac{145}{96\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}+{\frac{29}{40\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}-{\frac{13}{64\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{13}{768\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{3}{512\, \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))^4,x)

[Out]

16/d/a^4*(-1/768/(tan(1/2*d*x+1/2*c)-1)^3-1/512/(tan(1/2*d*x+1/2*c)-1)^2-1/512/(tan(1/2*d*x+1/2*c)-1)+1/11/(ta
n(1/2*d*x+1/2*c)+1)^11-1/2/(tan(1/2*d*x+1/2*c)+1)^10+23/18/(tan(1/2*d*x+1/2*c)+1)^9-2/(tan(1/2*d*x+1/2*c)+1)^8
+235/112/(tan(1/2*d*x+1/2*c)+1)^7-145/96/(tan(1/2*d*x+1/2*c)+1)^6+29/40/(tan(1/2*d*x+1/2*c)+1)^5-13/64/(tan(1/
2*d*x+1/2*c)+1)^4+13/768/(tan(1/2*d*x+1/2*c)+1)^3+3/512/(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 = 1.15319, size = 686, normalized size = 4.73 \begin{align*} \text{result too large to display} \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))^4,x, algorithm="maxima")

[Out]

4/3465*(488*sin(d*x + c)/(cos(d*x + c) + 1) + 1525*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1952*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 + 2794*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 176*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 481
8*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5280*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 10857*sin(d*x + c)^8/(cos(d
*x + c) + 1)^8 + 5544*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 3465*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 61)/(
(a^4 + 8*a^4*sin(d*x + c)/(cos(d*x + c) + 1) + 25*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 32*a^4*sin(d*x + c
)^3/(cos(d*x + c) + 1)^3 - 11*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 88*a^4*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 - 99*a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 99*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 88*a^4*sin(d*
x + c)^9/(cos(d*x + c) + 1)^9 + 11*a^4*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 32*a^4*sin(d*x + c)^11/(cos(d*x
+ c) + 1)^11 - 25*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^4*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - a
^4*sin(d*x + c)^14/(cos(d*x + c) + 1)^14)*d)

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Fricas [A]  time = 1.74186, size = 402, normalized size = 2.77 \begin{align*} -\frac{128 \, \cos \left (d x + c\right )^{6} - 320 \, \cos \left (d x + c\right )^{4} + 805 \, \cos \left (d x + c\right )^{2} + 4 \,{\left (8 \, \cos \left (d x + c\right )^{6} - 60 \, \cos \left (d x + c\right )^{4} + 35 \, \cos \left (d x + c\right )^{2} - 105\right )} \sin \left (d x + c\right ) - 735}{3465 \,{\left (a^{4} d \cos \left (d x + c\right )^{7} - 8 \, a^{4} d \cos \left (d x + c\right )^{5} + 8 \, a^{4} d \cos \left (d x + c\right )^{3} - 4 \,{\left (a^{4} d \cos \left (d x + c\right )^{5} - 2 \, a^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/3465*(128*cos(d*x + c)^6 - 320*cos(d*x + c)^4 + 805*cos(d*x + c)^2 + 4*(8*cos(d*x + c)^6 - 60*cos(d*x + c)^
4 + 35*cos(d*x + c)^2 - 105)*sin(d*x + c) - 735)/(a^4*d*cos(d*x + c)^7 - 8*a^4*d*cos(d*x + c)^5 + 8*a^4*d*cos(
d*x + c)^3 - 4*(a^4*d*cos(d*x + c)^5 - 2*a^4*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))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.33721, size = 267, normalized size = 1.84 \begin{align*} -\frac{\frac{1155 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2\right )}}{a^{4}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{3465 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 45045 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 279510 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 669900 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1205358 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 1334718 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1144440 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 627660 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 257345 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 57013 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5498}{a^{4}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{11}}}{110880 \, 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))^4,x, algorithm="giac")

[Out]

-1/110880*(1155*(3*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 2)/(a^4*(tan(1/2*d*x + 1/2*c) - 1)^3) - (
3465*tan(1/2*d*x + 1/2*c)^10 + 45045*tan(1/2*d*x + 1/2*c)^9 + 279510*tan(1/2*d*x + 1/2*c)^8 + 669900*tan(1/2*d
*x + 1/2*c)^7 + 1205358*tan(1/2*d*x + 1/2*c)^6 + 1334718*tan(1/2*d*x + 1/2*c)^5 + 1144440*tan(1/2*d*x + 1/2*c)
^4 + 627660*tan(1/2*d*x + 1/2*c)^3 + 257345*tan(1/2*d*x + 1/2*c)^2 + 57013*tan(1/2*d*x + 1/2*c) + 5498)/(a^4*(
tan(1/2*d*x + 1/2*c) + 1)^11))/d