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

Optimal. Leaf size=127 \[ \frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{5 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{9 \sec ^7(c+d x)}{7 a^3 d}-\frac{6 \sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d} \]

[Out]

Sec[c + d*x]^3/(3*a^3*d) - (6*Sec[c + d*x]^5)/(5*a^3*d) + (9*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9
*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) + (5*Tan[c + d*x]^7)/(7*a^3*d) + (4*Tan[c + d*x]^9)/(9*a^3*d)

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Rubi [A]  time = 0.223219, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2711, 2607, 270, 2606, 14} \[ \frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{5 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{9 \sec ^7(c+d x)}{7 a^3 d}-\frac{6 \sec ^5(c+d x)}{5 a^3 d}+\frac{\sec ^3(c+d x)}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sec[c + d*x]^3/(3*a^3*d) - (6*Sec[c + d*x]^5)/(5*a^3*d) + (9*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9
*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) + (5*Tan[c + d*x]^7)/(7*a^3*d) + (4*Tan[c + d*x]^9)/(9*a^3*d)

Rule 2711

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 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 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 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]]

Rubi steps

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

Mathematica [A]  time = 0.279077, size = 185, normalized size = 1.46 \[ \frac{39168 \sin (c+d x)+837 \sin (2 (c+d x))-28288 \sin (3 (c+d x))+372 \sin (4 (c+d x))+4224 \sin (5 (c+d x))-31 \sin (6 (c+d x))+1116 \cos (c+d x)-21312 \cos (2 (c+d x))+62 \cos (3 (c+d x))+8448 \cos (4 (c+d x))-186 \cos (5 (c+d x))-704 \cos (6 (c+d x))+5376}{322560 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5376 + 1116*Cos[c + d*x] - 21312*Cos[2*(c + d*x)] + 62*Cos[3*(c + d*x)] + 8448*Cos[4*(c + d*x)] - 186*Cos[5*(
c + d*x)] - 704*Cos[6*(c + d*x)] + 39168*Sin[c + d*x] + 837*Sin[2*(c + d*x)] - 28288*Sin[3*(c + d*x)] + 372*Si
n[4*(c + d*x)] + 4224*Sin[5*(c + d*x)] - 31*Sin[6*(c + d*x)])/(322560*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 = 0.141, size = 175, normalized size = 1.4 \begin{align*} 32\,{\frac{1}{d{a}^{3}} \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}{1024\,\tan \left ( 1/2\,dx+c/2 \right ) -1024}}-1/36\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-9}+1/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}-{\frac{13}{56\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{7}}}+{\frac{11}{48\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}-{\frac{39}{320\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+{\frac{3}{128\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{1}{192\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-{\frac{1}{1024\,\tan \left ( 1/2\,dx+c/2 \right ) +1024}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/d/a^3*(-1/768/(tan(1/2*d*x+1/2*c)-1)^3-1/512/(tan(1/2*d*x+1/2*c)-1)^2+1/1024/(tan(1/2*d*x+1/2*c)-1)-1/36/(t
an(1/2*d*x+1/2*c)+1)^9+1/8/(tan(1/2*d*x+1/2*c)+1)^8-13/56/(tan(1/2*d*x+1/2*c)+1)^7+11/48/(tan(1/2*d*x+1/2*c)+1
)^6-39/320/(tan(1/2*d*x+1/2*c)+1)^5+3/128/(tan(1/2*d*x+1/2*c)+1)^4+1/192/(tan(1/2*d*x+1/2*c)+1)^3-1/1024/(tan(
1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.06783, size = 543, normalized size = 4.28 \begin{align*} -\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{162 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{126 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{126 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 1\right )}}{315 \,{\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)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-16/315*(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 - 162*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 126*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 126*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 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 = 1.7035, size = 335, normalized size = 2.64 \begin{align*} \frac{22 \, \cos \left (d x + c\right )^{6} - 99 \, \cos \left (d x + c\right )^{4} + 120 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (33 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) - 35}{315 \,{\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)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/315*(22*cos(d*x + c)^6 - 99*cos(d*x + c)^4 + 120*cos(d*x + c)^2 - 2*(33*cos(d*x + c)^4 - 80*cos(d*x + c)^2 +
35)*sin(d*x + c) - 35)/(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)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25984, size = 215, normalized size = 1.69 \begin{align*} \frac{\frac{105 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{315 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 2520 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 7140 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1638 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8232 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2988 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 432 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 13}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10080*(105*(3*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 5)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (31
5*tan(1/2*d*x + 1/2*c)^8 + 2520*tan(1/2*d*x + 1/2*c)^7 + 7140*tan(1/2*d*x + 1/2*c)^6 - 1638*tan(1/2*d*x + 1/2*
c)^4 - 8232*tan(1/2*d*x + 1/2*c)^3 - 2988*tan(1/2*d*x + 1/2*c)^2 - 432*tan(1/2*d*x + 1/2*c) - 13)/(a^3*(tan(1/
2*d*x + 1/2*c) + 1)^9))/d