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

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

[Out]

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

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Rubi [A]  time = 0.138481, 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, 14, 2607, 30} \[ -\frac{\tan ^5(c+d x)}{5 a d}+\frac{\sec ^5(c+d x)}{5 a d}-\frac{\sec ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[c + d*x]^3/(3*a*d) + Sec[c + d*x]^5/(5*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 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 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 30

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

Rubi steps

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

Mathematica [A]  time = 0.288426, size = 106, normalized size = 1.93 \[ \frac{\sec ^3(c+d x) (16 \sin (c+d x)+22 \sin (2 (c+d x))-48 \sin (3 (c+d x))+11 \sin (4 (c+d x))+66 \cos (c+d x)-192 \cos (2 (c+d x))+22 \cos (3 (c+d x))+24 \cos (4 (c+d x))+40)}{960 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(40 + 66*Cos[c + d*x] - 192*Cos[2*(c + d*x)] + 22*Cos[3*(c + d*x)] + 24*Cos[4*(c + d*x)] + 16*
Sin[c + d*x] + 22*Sin[2*(c + d*x)] - 48*Sin[3*(c + d*x)] + 11*Sin[4*(c + d*x)]))/(960*a*d*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.081, size = 115, normalized size = 2.1 \begin{align*} 16\,{\frac{1}{da} \left ( -{\frac{1}{96\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{3}}}-{\frac{1}{64\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}+{\frac{1}{128\,\tan \left ( 1/2\,dx+c/2 \right ) -128}}+1/40\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}-1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+1/24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}-{\frac{1}{128\,\tan \left ( 1/2\,dx+c/2 \right ) +128}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/d/a*(-1/96/(tan(1/2*d*x+1/2*c)-1)^3-1/64/(tan(1/2*d*x+1/2*c)-1)^2+1/128/(tan(1/2*d*x+1/2*c)-1)+1/40/(tan(1/
2*d*x+1/2*c)+1)^5-1/16/(tan(1/2*d*x+1/2*c)+1)^4+1/24/(tan(1/2*d*x+1/2*c)+1)^3-1/128/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.06878, size = 316, normalized size = 5.75 \begin{align*} -\frac{4 \,{\left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{6 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\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)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-4/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)/((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/(c
os(d*x + c) + 1)^8)*d)

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Fricas [A]  time = 1.34623, size = 190, normalized size = 3.45 \begin{align*} \frac{3 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \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)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/15*(3*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - (3*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)

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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)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.24477, size = 162, normalized size = 2.95 \begin{align*} \frac{\frac{5 \,{\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{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{15 \, \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} + 10 \, \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 ) + 7}{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)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/120*(5*(3*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 5)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) - (15*tan(1
/2*d*x + 1/2*c)^4 + 60*tan(1/2*d*x + 1/2*c)^3 + 10*tan(1/2*d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) + 7)/(a*(t
an(1/2*d*x + 1/2*c) + 1)^5))/d