3.808 \(\int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=73 \[ \frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

-((a^2*ArcTanh[Cos[c + d*x]])/d) + (4*a^2*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x])/(3*d*(a
- a*Sin[c + d*x])^2)

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Rubi [A]  time = 0.187461, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2869, 2766, 2978, 12, 3770} \[ \frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

-((a^2*ArcTanh[Cos[c + d*x]])/d) + (4*a^2*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x])/(3*d*(a
- a*Sin[c + d*x])^2)

Rule 2869

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

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx &=a^4 \int \frac{\csc (c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} a^2 \int \frac{\csc (c+d x) (3 a+a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=\frac{4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} \int 3 a^2 \csc (c+d x) \, dx\\ &=\frac{4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+a^2 \int \csc (c+d x) \, dx\\ &=-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.501146, size = 142, normalized size = 1.95 \[ \frac{a^2 (\sin (c+d x)+1)^2 \left (3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (4 \sin (c+d x)-5)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{1}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.103, size = 92, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{2}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{4\,{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \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)^4*(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.02628, size = 122, normalized size = 1.67 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{2 \, a^{2}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + a^2*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x +
c) + 1) + 3*log(cos(d*x + c) - 1)) + 2*a^2/cos(d*x + c)^3)/d

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Fricas [B]  time = 1.41095, size = 578, normalized size = 7.92 \begin{align*} -\frac{8 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 2 \, a^{2} + 3 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} +{\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} +{\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (4 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right ) + 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)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(8*a^2*cos(d*x + c)^2 + 10*a^2*cos(d*x + c) + 2*a^2 + 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2 +
(a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c
) - 2*a^2 + (a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(4*a^2*cos(d*x + c) - a^
2)*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 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)*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.26965, size = 99, normalized size = 1.36 \begin{align*} \frac{3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(3*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(6*a^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*tan(1/2*d*x + 1/2*c) + 5*a
^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d