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

Optimal. Leaf size=105 \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {11 \sec ^7(c+d x)}{7 a^3 d}+\frac {2 \sec ^5(c+d x)}{a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d} \]

[Out]

-sec(d*x+c)^3/a^3/d+2*sec(d*x+c)^5/a^3/d-11/7*sec(d*x+c)^7/a^3/d+4/9*sec(d*x+c)^9/a^3/d-3/7*tan(d*x+c)^7/a^3/d
-4/9*tan(d*x+c)^9/a^3/d

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Rubi [A]  time = 0.34, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2875, 2873, 2606, 270, 2607, 14, 30} \[ -\frac {4 \tan ^9(c+d x)}{9 a^3 d}-\frac {3 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \sec ^9(c+d x)}{9 a^3 d}-\frac {11 \sec ^7(c+d x)}{7 a^3 d}+\frac {2 \sec ^5(c+d x)}{a^3 d}-\frac {\sec ^3(c+d x)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3/(a^3*d)) + (2*Sec[c + d*x]^5)/(a^3*d) - (11*Sec[c + d*x]^7)/(7*a^3*d) + (4*Sec[c + d*x]^9)/(9
*a^3*d) - (3*Tan[c + d*x]^7)/(7*a^3*d) - (4*Tan[c + d*x]^9)/(9*a^3*d)

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 185, normalized size = 1.76 \[ \frac {-1152 \sin (c+d x)+6507 \sin (2 (c+d x))-8128 \sin (3 (c+d x))+2892 \sin (4 (c+d x))+192 \sin (5 (c+d x))-241 \sin (6 (c+d x))+8676 \cos (c+d x)-11232 \cos (2 (c+d x))+482 \cos (3 (c+d x))+4416 \cos (4 (c+d x))-1446 \cos (5 (c+d x))-32 \cos (6 (c+d x))-1344}{64512 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1344 + 8676*Cos[c + d*x] - 11232*Cos[2*(c + d*x)] + 482*Cos[3*(c + d*x)] + 4416*Cos[4*(c + d*x)] - 1446*Cos[
5*(c + d*x)] - 32*Cos[6*(c + d*x)] - 1152*Sin[c + d*x] + 6507*Sin[2*(c + d*x)] - 8128*Sin[3*(c + d*x)] + 2892*
Sin[4*(c + d*x)] + 192*Sin[5*(c + d*x)] - 241*Sin[6*(c + d*x)])/(64512*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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fricas [A]  time = 0.47, size = 128, normalized size = 1.22 \[ \frac {\cos \left (d x + c\right )^{6} - 36 \, \cos \left (d x + c\right )^{4} + 57 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{4} - 34 \, \cos \left (d x + c\right )^{2} + 7\right )} \sin \left (d x + c\right ) - 14}{63 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/63*(cos(d*x + c)^6 - 36*cos(d*x + c)^4 + 57*cos(d*x + c)^2 - (3*cos(d*x + c)^4 - 34*cos(d*x + c)^2 + 7)*sin(
d*x + c) - 14)/(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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giac [A]  time = 0.39, size = 172, normalized size = 1.64 \[ \frac {\frac {21 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {189 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 1764 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7224 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 16380 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 19026 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16380 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8352 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2340 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 281}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{2016 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2016*(21*(9*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 11)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (189
*tan(1/2*d*x + 1/2*c)^8 + 1764*tan(1/2*d*x + 1/2*c)^7 + 7224*tan(1/2*d*x + 1/2*c)^6 + 16380*tan(1/2*d*x + 1/2*
c)^5 + 19026*tan(1/2*d*x + 1/2*c)^4 + 16380*tan(1/2*d*x + 1/2*c)^3 + 8352*tan(1/2*d*x + 1/2*c)^2 + 2340*tan(1/
2*d*x + 1/2*c) + 281)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9))/d

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maple [A]  time = 0.56, size = 190, normalized size = 1.81 \[ \frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}+\frac {48}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x)

[Out]

64/d/a^3*(-1/1536/(tan(1/2*d*x+1/2*c)-1)^3-1/1024/(tan(1/2*d*x+1/2*c)-1)^2+3/2048/(tan(1/2*d*x+1/2*c)-1)+1/72/
(tan(1/2*d*x+1/2*c)+1)^9-1/16/(tan(1/2*d*x+1/2*c)+1)^8+3/28/(tan(1/2*d*x+1/2*c)+1)^7-1/12/(tan(1/2*d*x+1/2*c)+
1)^6+3/128/(tan(1/2*d*x+1/2*c)+1)^5+1/256/(tan(1/2*d*x+1/2*c)+1)^4-1/768/(tan(1/2*d*x+1/2*c)+1)^3-1/512/(tan(1
/2*d*x+1/2*c)+1)^2-3/2048/(tan(1/2*d*x+1/2*c)+1))

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maxima [B]  time = 0.34, size = 382, normalized size = 3.64 \[ -\frac {16 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {42 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}{63 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-16/63*(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 - 27*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 42*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + 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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mupad [B]  time = 13.19, size = 208, normalized size = 1.98 \[ -\frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{63}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}+\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{21}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{63}-\frac {48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}-\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{7}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(cos(c + d*x)^4*(a + a*sin(c + d*x))^3),x)

[Out]

-((16*cos(c/2 + (d*x)/2)^12)/63 + (32*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2))/21 - (32*cos(c/2 + (d*x)/2)^6*
sin(c/2 + (d*x)/2)^6)/3 - (64*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5)/7 - (48*cos(c/2 + (d*x)/2)^8*sin(c/2
+ (d*x)/2)^4)/7 + (32*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3)/63 + (64*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x
)/2)^2)/21)/(a^3*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)

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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)**4*sin(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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