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

Optimal. Leaf size=103 \[ \frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{a^3 d}-\frac{\sec ^3(c+d x)}{3 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.243329, antiderivative size = 103, 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} \[ \frac{4 \tan ^7(c+d x)}{7 a^3 d}+\frac{\tan ^5(c+d x)}{a^3 d}+\frac{\tan ^3(c+d x)}{3 a^3 d}-\frac{4 \sec ^7(c+d x)}{7 a^3 d}+\frac{\sec ^5(c+d x)}{a^3 d}-\frac{\sec ^3(c+d x)}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.338808, size = 104, normalized size = 1.01 \[ \frac{\sec (c+d x) (672 \sin (c+d x)-70 \sin (2 (c+d x))-96 \sin (3 (c+d x))+5 \sin (4 (c+d x))-70 \cos (c+d x)-224 \cos (2 (c+d x))+30 \cos (3 (c+d x))+16 \cos (4 (c+d x))+336)}{1344 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(336 - 70*Cos[c + d*x] - 224*Cos[2*(c + d*x)] + 30*Cos[3*(c + d*x)] + 16*Cos[4*(c + d*x)] + 672*
Sin[c + d*x] - 70*Sin[2*(c + d*x)] - 96*Sin[3*(c + d*x)] + 5*Sin[4*(c + d*x)]))/(1344*a^3*d*(1 + Sin[c + d*x])
^3)

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Maple [A]  time = 0.104, size = 130, normalized size = 1.3 \begin{align*} 8\,{\frac{1}{d{a}^{3}} \left ( -{\frac{1}{64\,\tan \left ( 1/2\,dx+c/2 \right ) -64}}-1/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-7}+1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-6}-3/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+5/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-{\frac{13}{48\, \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{1}{64\,\tan \left ( 1/2\,dx+c/2 \right ) +64}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/d/a^3*(-1/64/(tan(1/2*d*x+1/2*c)-1)-1/7/(tan(1/2*d*x+1/2*c)+1)^7+1/2/(tan(1/2*d*x+1/2*c)+1)^6-3/4/(tan(1/2*d
*x+1/2*c)+1)^5+5/8/(tan(1/2*d*x+1/2*c)+1)^4-13/48/(tan(1/2*d*x+1/2*c)+1)^3+1/32/(tan(1/2*d*x+1/2*c)+1)^2+1/64/
(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.04846, size = 365, normalized size = 3.54 \begin{align*} \frac{4 \,{\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{28 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{14 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{21 \,{\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)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

4/21*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 14*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 21*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 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 = 1.14163, size = 265, normalized size = 2.57 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - 6 \,{\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 9}{21 \,{\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)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/21*(2*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - 6*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 9)/(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)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.24318, size = 162, normalized size = 1.57 \begin{align*} -\frac{\frac{21}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} - \frac{21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 168 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 161 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 224 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 63 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 56 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{7}}}{168 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*(21/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (21*tan(1/2*d*x + 1/2*c)^6 + 168*tan(1/2*d*x + 1/2*c)^5 + 161*ta
n(1/2*d*x + 1/2*c)^4 + 224*tan(1/2*d*x + 1/2*c)^3 + 63*tan(1/2*d*x + 1/2*c)^2 + 56*tan(1/2*d*x + 1/2*c) + 11)/
(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d