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

Optimal. Leaf size=117 \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{11 a \log (1-\sin (c+d x))}{16 d}+\frac{a \log (\sin (c+d x))}{d}-\frac{5 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

(-11*a*Log[1 - Sin[c + d*x]])/(16*d) + (a*Log[Sin[c + d*x]])/d - (5*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d
*(a - a*Sin[c + d*x])^2) + a^2/(2*d*(a - a*Sin[c + d*x])) + a^2/(8*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.107274, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2836, 12, 88} \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{11 a \log (1-\sin (c+d x))}{16 d}+\frac{a \log (\sin (c+d x))}{d}-\frac{5 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*a*Log[1 - Sin[c + d*x]])/(16*d) + (a*Log[Sin[c + d*x]])/d - (5*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d
*(a - a*Sin[c + d*x])^2) + a^2/(2*d*(a - a*Sin[c + d*x])) + a^2/(8*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{a}{(a-x)^3 x (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 x (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \left (\frac{1}{4 a^3 (a-x)^3}+\frac{1}{2 a^4 (a-x)^2}+\frac{11}{16 a^5 (a-x)}+\frac{1}{a^5 x}-\frac{1}{8 a^4 (a+x)^2}-\frac{5}{16 a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{11 a \log (1-\sin (c+d x))}{16 d}+\frac{a \log (\sin (c+d x))}{d}-\frac{5 a \log (1+\sin (c+d x))}{16 d}+\frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.209148, size = 99, normalized size = 0.85 \[ \frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{a \left (-\sec ^4(c+d x)-2 \sec ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d}+\frac{3 a \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*Log[Cos[c + d*x]] - 4*Log[Sin[c + d*x]] - 2*Sec[c + d*x]^2 - Sec[c + d*x]^4))/(4*d) + (a*Sec[c + d*x]^3
*Tan[c + d*x])/(4*d) + (3*a*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x]))/(8*d)

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Maple [A]  time = 0.085, size = 100, normalized size = 0.9 \begin{align*}{\frac{a\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,a\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{a}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{a}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*a*tan(d*x+c)*sec(d*x+c)^3+3/8*a*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*a/cos(d*
x+c)^4+1/2/d*a/cos(d*x+c)^2+a*ln(tan(d*x+c))/d

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Maxima [A]  time = 1.01337, size = 128, normalized size = 1.09 \begin{align*} -\frac{5 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac{2 \,{\left (3 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(5*a*log(sin(d*x + c) + 1) + 11*a*log(sin(d*x + c) - 1) - 16*a*log(sin(d*x + c)) + 2*(3*a*sin(d*x + c)^2
+ a*sin(d*x + c) - 6*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.44537, size = 456, normalized size = 3.9 \begin{align*} -\frac{6 \, a \cos \left (d x + c\right )^{2} - 16 \,{\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 5 \,{\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \,{\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a \sin \left (d x + c\right ) + 6 \, a}{16 \,{\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*a*cos(d*x + c)^2 - 16*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(1/2*sin(d*x + c)) + 5*(a
*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 11*(a*cos(d*x + c)^2*sin(d*x + c) - a
*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*a*sin(d*x + c) + 6*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x +
c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.32802, size = 140, normalized size = 1.2 \begin{align*} -\frac{10 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 22 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 32 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{2 \,{\left (5 \, a \sin \left (d x + c\right ) + 7 \, a\right )}}{\sin \left (d x + c\right ) + 1} - \frac{33 \, a \sin \left (d x + c\right )^{2} - 82 \, a \sin \left (d x + c\right ) + 53 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(10*a*log(abs(sin(d*x + c) + 1)) + 22*a*log(abs(sin(d*x + c) - 1)) - 32*a*log(abs(sin(d*x + c))) - 2*(5*
a*sin(d*x + c) + 7*a)/(sin(d*x + c) + 1) - (33*a*sin(d*x + c)^2 - 82*a*sin(d*x + c) + 53*a)/(sin(d*x + c) - 1)
^2)/d