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

Optimal. Leaf size=55 \[ -\frac{\tan ^5(c+d x)}{5 a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\sec ^5(c+d x)}{5 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107935, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2839, 2606, 30, 2607, 14} \[ -\frac{\tan ^5(c+d x)}{5 a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\sec ^5(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \sec ^5(c+d x) \tan (c+d x) \, dx}{a}-\frac{\int \sec ^4(c+d x) \tan ^2(c+d x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{a d}-\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{\sec ^5(c+d x)}{5 a d}-\frac{\operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{\sec ^5(c+d x)}{5 a d}-\frac{\tan ^3(c+d x)}{3 a d}-\frac{\tan ^5(c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.259062, size = 106, normalized size = 1.93 \[ -\frac{\sec ^3(c+d x) (-96 \sin (c+d x)+18 \sin (2 (c+d x))-32 \sin (3 (c+d x))+9 \sin (4 (c+d x))+54 \cos (c+d x)+32 \cos (2 (c+d x))+18 \cos (3 (c+d x))+16 \cos (4 (c+d x))-240)}{960 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(-240 + 54*Cos[c + d*x] + 32*Cos[2*(c + d*x)] + 18*Cos[3*(c + d*x)] + 16*Cos[4*(c + d*x)] - 9
6*Sin[c + d*x] + 18*Sin[2*(c + d*x)] - 32*Sin[3*(c + d*x)] + 9*Sin[4*(c + d*x)]))/(960*a*d*(1 + Sin[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.067, size = 130, normalized size = 2.4 \begin{align*} 4\,{\frac{1}{da} \left ( -1/24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-3}-1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-2}-{\frac{3}{32\,\tan \left ( 1/2\,dx+c/2 \right ) -32}}-1/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+1/10\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+1/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}-1/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+{\frac{3}{32\,\tan \left ( 1/2\,dx+c/2 \right ) +32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/a*(-1/24/(tan(1/2*d*x+1/2*c)-1)^3-1/16/(tan(1/2*d*x+1/2*c)-1)^2-3/32/(tan(1/2*d*x+1/2*c)-1)-1/4/(tan(1/2*d
*x+1/2*c)+1)^4+1/10/(tan(1/2*d*x+1/2*c)+1)^5+1/3/(tan(1/2*d*x+1/2*c)+1)^3-1/4/(tan(1/2*d*x+1/2*c)+1)^2+3/32/(t
an(1/2*d*x+1/2*c)+1))

________________________________________________________________________________________

Maxima [B]  time = 1.18115, size = 370, normalized size = 6.73 \begin{align*} \frac{2 \,{\left (\frac{6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{10 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 3\right )}}{15 \,{\left (a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a \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)^4*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

2/15*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^6
/(cos(d*x + c) + 1)^6 + 3)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos
(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

________________________________________________________________________________________

Fricas [A]  time = 1.31488, size = 189, normalized size = 3.44 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} -{\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 4}{15 \,{\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(2*cos(d*x + c)^4 - cos(d*x + c)^2 - (2*cos(d*x + c)^2 + 1)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^3*sin(d*
x + c) + a*d*cos(d*x + c)^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sin{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

Integral(sin(c + d*x)*sec(c + d*x)**4/(sin(c + d*x) + 1), x)/a

________________________________________________________________________________________

Giac [B]  time = 1.21636, size = 162, normalized size = 2.95 \begin{align*} -\frac{\frac{5 \,{\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{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 60 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 70 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 13}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(5*(9*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 7)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) - (45*tan(
1/2*d*x + 1/2*c)^4 + 60*tan(1/2*d*x + 1/2*c)^3 + 70*tan(1/2*d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) + 13)/(a*
(tan(1/2*d*x + 1/2*c) + 1)^5))/d