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

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

[Out]

-Sec[c + d*x]^5/(5*a^3*d) + (5*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + Tan[c + d*x]^3/(3*a^
3*d) + (6*Tan[c + d*x]^5)/(5*a^3*d) + (9*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 = 0.362324, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2875, 2873, 2607, 270, 2606, 14} \[ \frac{4 \tan ^9(c+d x)}{9 a^3 d}+\frac{9 \tan ^7(c+d x)}{7 a^3 d}+\frac{6 \tan ^5(c+d x)}{5 a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{5 \sec ^7(c+d x)}{7 a^3 d}-\frac{\sec ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[c + d*x]^5/(5*a^3*d) + (5*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + Tan[c + d*x]^3/(3*a^
3*d) + (6*Tan[c + d*x]^5)/(5*a^3*d) + (9*Tan[c + d*x]^7)/(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 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{\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \sec ^8(c+d x) (a-a \sin (c+d x))^3 \tan ^2(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (a^3 \sec ^8(c+d x) \tan ^2(c+d x)-3 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)-a^3 \sec ^5(c+d x) \tan ^5(c+d x)\right ) \, dx}{a^6}\\ &=\frac{\int \sec ^8(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac{\int \sec ^5(c+d x) \tan ^5(c+d x) \, dx}{a^3}-\frac{3 \int \sec ^7(c+d x) \tan ^3(c+d x) \, dx}{a^3}+\frac{3 \int \sec ^6(c+d x) \tan ^4(c+d x) \, dx}{a^3}\\ &=-\frac{\operatorname{Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^6 \left (-1+x^2\right ) \, 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^4-2 x^6+x^8\right ) \, dx,x,\sec (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,\tan (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (-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{\sec ^5(c+d x)}{5 a^3 d}+\frac{5 \sec ^7(c+d x)}{7 a^3 d}-\frac{4 \sec ^9(c+d x)}{9 a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}+\frac{6 \tan ^5(c+d x)}{5 a^3 d}+\frac{9 \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 = 0.328035, size = 185, normalized size = 1.46 \[ \frac{73728 \sin (c+d x)-7263 \sin (2 (c+d x))+512 \sin (3 (c+d x))-3228 \sin (4 (c+d x))-1536 \sin (5 (c+d x))+269 \sin (6 (c+d x))-9684 \cos (c+d x)-6912 \cos (2 (c+d x))-538 \cos (3 (c+d x))-3072 \cos (4 (c+d x))+1614 \cos (5 (c+d x))+256 \cos (6 (c+d x))+32256}{322560 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32256 - 9684*Cos[c + d*x] - 6912*Cos[2*(c + d*x)] - 538*Cos[3*(c + d*x)] - 3072*Cos[4*(c + d*x)] + 1614*Cos[5
*(c + d*x)] + 256*Cos[6*(c + d*x)] + 73728*Sin[c + d*x] - 7263*Sin[2*(c + d*x)] + 512*Sin[3*(c + d*x)] - 3228*
Sin[4*(c + d*x)] - 1536*Sin[5*(c + d*x)] + 269*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 = 0.125, size = 190, normalized size = 1.5 \begin{align*} 8\,{\frac{1}{d{a}^{3}} \left ( -{\frac{1}{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{3}{256\,\tan \left ( 1/2\,dx+c/2 \right ) -256}}-1/9\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-9}+1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}-{\frac{15}{14\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{7}}}+{\frac{17}{12\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}-{\frac{99}{80\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+{\frac{23}{32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-1/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}+1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+{\frac{3}{256\,\tan \left ( 1/2\,dx+c/2 \right ) +256}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x)

[Out]

8/d/a^3*(-1/192/(tan(1/2*d*x+1/2*c)-1)^3-1/128/(tan(1/2*d*x+1/2*c)-1)^2-3/256/(tan(1/2*d*x+1/2*c)-1)-1/9/(tan(
1/2*d*x+1/2*c)+1)^9+1/2/(tan(1/2*d*x+1/2*c)+1)^8-15/14/(tan(1/2*d*x+1/2*c)+1)^7+17/12/(tan(1/2*d*x+1/2*c)+1)^6
-99/80/(tan(1/2*d*x+1/2*c)+1)^5+23/32/(tan(1/2*d*x+1/2*c)+1)^4-1/4/(tan(1/2*d*x+1/2*c)+1)^3+1/32/(tan(1/2*d*x+
1/2*c)+1)^2+3/256/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.16055, size = 597, normalized size = 4.7 \begin{align*} \frac{4 \,{\left (\frac{66 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{232 \, \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{108 \, \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{504 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{315 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{210 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 11\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)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

4/315*(66*sin(d*x + c)/(cos(d*x + c) + 1) + 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 232*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 18*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 108*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 84*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6 + 504*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 315*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 + 210*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 11)/((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 = 1.6539, size = 333, normalized size = 2.62 \begin{align*} -\frac{8 \, \cos \left (d x + c\right )^{6} - 36 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} - 2 \,{\left (12 \, \cos \left (d x + c\right )^{4} - 10 \, \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)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/315*(8*cos(d*x + c)^6 - 36*cos(d*x + c)^4 + 15*cos(d*x + c)^2 - 2*(12*cos(d*x + c)^4 - 10*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)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29245, size = 232, normalized size = 1.83 \begin{align*} -\frac{\frac{105 \,{\left (9 \, \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 ) + 7\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{945 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 10080 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 23940 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 42840 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 41958 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 32592 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 14148 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5112 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 673}{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)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/10080*(105*(9*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 7)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (9
45*tan(1/2*d*x + 1/2*c)^8 + 10080*tan(1/2*d*x + 1/2*c)^7 + 23940*tan(1/2*d*x + 1/2*c)^6 + 42840*tan(1/2*d*x +
1/2*c)^5 + 41958*tan(1/2*d*x + 1/2*c)^4 + 32592*tan(1/2*d*x + 1/2*c)^3 + 14148*tan(1/2*d*x + 1/2*c)^2 + 5112*t
an(1/2*d*x + 1/2*c) + 673)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9))/d