3.876 \(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=93 \[ \frac{a^5}{2 d (a-a \sin (c+d x))^2}+\frac{2 a^4}{d (a-a \sin (c+d x))}-\frac{a^3 \csc (c+d x)}{d}-\frac{3 a^3 \log (1-\sin (c+d x))}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

[Out]

-((a^3*Csc[c + d*x])/d) - (3*a^3*Log[1 - Sin[c + d*x]])/d + (3*a^3*Log[Sin[c + d*x]])/d + a^5/(2*d*(a - a*Sin[
c + d*x])^2) + (2*a^4)/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.157522, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 44} \[ \frac{a^5}{2 d (a-a \sin (c+d x))^2}+\frac{2 a^4}{d (a-a \sin (c+d x))}-\frac{a^3 \csc (c+d x)}{d}-\frac{3 a^3 \log (1-\sin (c+d x))}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

-((a^3*Csc[c + d*x])/d) - (3*a^3*Log[1 - Sin[c + d*x]])/d + (3*a^3*Log[Sin[c + d*x]])/d + a^5/(2*d*(a - a*Sin[
c + d*x])^2) + (2*a^4)/(d*(a - a*Sin[c + d*x]))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.221222, size = 63, normalized size = 0.68 \[ \frac{a^3 \left (-\frac{4}{\sin (c+d x)-1}+\frac{1}{(\sin (c+d x)-1)^2}-2 \csc (c+d x)-6 \log (1-\sin (c+d x))+6 \log (\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(-2*Csc[c + d*x] - 6*Log[1 - Sin[c + d*x]] + 6*Log[Sin[c + d*x]] + (-1 + Sin[c + d*x])^(-2) - 4/(-1 + Sin
[c + d*x])))/(2*d)

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Maple [A]  time = 0.136, size = 176, normalized size = 1.9 \begin{align*}{\frac{{a}^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+3\,{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{{a}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,{a}^{3}}{8\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15\,{a}^{3}}{8\,d\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/d*a^3/cos(d*x+c)^4+3/4/d*a^3*tan(d*x+c)*sec(d*x+c)^3+9/8/d*a^3*sec(d*x+c)*tan(d*x+c)+3/d*a^3*ln(sec(d*x+c)+t
an(d*x+c))+3/2/d*a^3/cos(d*x+c)^2+3/d*a^3*ln(tan(d*x+c))+1/4/d*a^3/sin(d*x+c)/cos(d*x+c)^4+5/8/d*a^3/sin(d*x+c
)/cos(d*x+c)^2-15/8/d*a^3/sin(d*x+c)

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Maxima [A]  time = 1.12938, size = 122, normalized size = 1.31 \begin{align*} -\frac{6 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac{6 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}}{\sin \left (d x + c\right )^{3} - 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(6*a^3*log(sin(d*x + c) - 1) - 6*a^3*log(sin(d*x + c)) + (6*a^3*sin(d*x + c)^2 - 9*a^3*sin(d*x + c) + 2*a
^3)/(sin(d*x + c)^3 - 2*sin(d*x + c)^2 + sin(d*x + c)))/d

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Fricas [A]  time = 1.4574, size = 441, normalized size = 4.74 \begin{align*} \frac{6 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) - 8 \, a^{3} + 6 \,{\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 6 \,{\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (2 \, d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right )^{2} - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^3*cos(d*x + c)^2 + 9*a^3*sin(d*x + c) - 8*a^3 + 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2
- 2*a^3)*sin(d*x + c))*log(1/2*sin(d*x + c)) - 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2 - 2*a^3)
*sin(d*x + c))*log(-sin(d*x + c) + 1))/(2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 2*d)*sin(d*x + c) - 2*d)

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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(csc(d*x+c)**2*sec(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20734, size = 224, normalized size = 2.41 \begin{align*} -\frac{12 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{25 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 88 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 130 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 88 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + a^3*tan(1/2*d*x + 1/2
*c) + (6*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) - (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 88*a^3*tan(1/
2*d*x + 1/2*c)^3 + 130*a^3*tan(1/2*d*x + 1/2*c)^2 - 88*a^3*tan(1/2*d*x + 1/2*c) + 25*a^3)/(tan(1/2*d*x + 1/2*c
) - 1)^4)/d