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

Optimal. Leaf size=102 \[ \frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{9 \sec ^5(c+d x)}{5 a^3 d}-\frac{2 \sec ^3(c+d x)}{a^3 d}+\frac{\sec (c+d x)}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.328677, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2875, 2873, 2607, 14, 2606, 270, 30, 194} \[ \frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{5 a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{9 \sec ^5(c+d x)}{5 a^3 d}-\frac{2 \sec ^3(c+d x)}{a^3 d}+\frac{\sec (c+d x)}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sec[c + d*x]/(a^3*d) - (2*Sec[c + d*x]^3)/(a^3*d) + (9*Sec[c + d*x]^5)/(5*a^3*d) - (4*Sec[c + d*x]^7)/(7*a^3*d
) + Tan[c + d*x]^5/(5*a^3*d) + (4*Tan[c + d*x]^7)/(7*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 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 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 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 30

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.477402, size = 104, normalized size = 1.02 \[ \frac{\sec (c+d x) (1344 \sin (c+d x)-1946 \sin (2 (c+d x))+64 \sin (3 (c+d x))+139 \sin (4 (c+d x))-1946 \cos (c+d x)-224 \cos (2 (c+d x))+834 \cos (3 (c+d x))-104 \cos (4 (c+d x))+840)}{2240 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(840 - 1946*Cos[c + d*x] - 224*Cos[2*(c + d*x)] + 834*Cos[3*(c + d*x)] - 104*Cos[4*(c + d*x)] +
1344*Sin[c + d*x] - 1946*Sin[2*(c + d*x)] + 64*Sin[3*(c + d*x)] + 139*Sin[4*(c + d*x)]))/(2240*a^3*d*(1 + Sin[
c + d*x])^3)

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Maple [A]  time = 0.107, size = 130, normalized size = 1.3 \begin{align*} 32\,{\frac{1}{d{a}^{3}} \left ( -{\frac{1}{256\,\tan \left ( 1/2\,dx+c/2 \right ) -256}}-1/28\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}+1/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-6}-{\frac{11}{80\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+{\frac{1}{64\, \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}{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)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

32/d/a^3*(-1/256/(tan(1/2*d*x+1/2*c)-1)-1/28/(tan(1/2*d*x+1/2*c)+1)^7+1/8/(tan(1/2*d*x+1/2*c)+1)^6-11/80/(tan(
1/2*d*x+1/2*c)+1)^5+1/32/(tan(1/2*d*x+1/2*c)+1)^4+1/64/(tan(1/2*d*x+1/2*c)+1)^3+1/128/(tan(1/2*d*x+1/2*c)+1)^2
+1/256/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.05044, size = 311, normalized size = 3.05 \begin{align*} \frac{16 \,{\left (\frac{6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{14 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{14 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{35 \,{\left (a^{3} + \frac{6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

16/35*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 14*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 1)/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
14*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)
^6/(cos(d*x + c) + 1)^6 - 6*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)
*d)

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Fricas [A]  time = 1.12336, size = 266, normalized size = 2.61 \begin{align*} \frac{13 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (\cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 15}{35 \,{\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) +{\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\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)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/35*(13*cos(d*x + c)^4 - 6*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + 5)*sin(d*x + c) - 15)/(3*a^3*d*cos(d*x + c)^3
- 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*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)**2*sin(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25186, size = 162, normalized size = 1.59 \begin{align*} -\frac{\frac{35}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} - \frac{35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 280 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1015 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1673 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 616 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 93}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (35*tan(1/2*d*x + 1/2*c)^6 + 280*tan(1/2*d*x + 1/2*c)^5 + 1015*t
an(1/2*d*x + 1/2*c)^4 + 2240*tan(1/2*d*x + 1/2*c)^3 + 1673*tan(1/2*d*x + 1/2*c)^2 + 616*tan(1/2*d*x + 1/2*c) +
93)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d