3.810 \(\int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ -\frac{16 a^2 \cot (c+d x)}{3 d}-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))} \]

[Out]

(-7*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (16*a^2*Cot[c + d*x])/(3*d) - (7*a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d) +
(8*a^2*Cot[c + d*x]*Csc[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cot[c + d*x]*Csc[c + d*x])/(3*d*(a - a*Sin[
c + d*x])^2)

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Rubi [A]  time = 0.30795, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2869, 2766, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{16 a^2 \cot (c+d x)}{3 d}-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(-7*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (16*a^2*Cot[c + d*x])/(3*d) - (7*a^2*Cot[c + d*x]*Csc[c + d*x])/(2*d) +
(8*a^2*Cot[c + d*x]*Csc[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cot[c + d*x]*Csc[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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \csc ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx &=a^4 \int \frac{\csc ^3(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} a^2 \int \frac{\csc ^3(c+d x) (5 a+3 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} \int \csc ^3(c+d x) \left (21 a^2+16 a^2 \sin (c+d x)\right ) \, dx\\ &=\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} \left (16 a^2\right ) \int \csc ^2(c+d x) \, dx+\left (7 a^2\right ) \int \csc ^3(c+d x) \, dx\\ &=-\frac{7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{2} \left (7 a^2\right ) \int \csc (c+d x) \, dx-\frac{\left (16 a^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d}\\ &=-\frac{7 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{16 a^2 \cot (c+d x)}{3 d}-\frac{7 a^2 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{8 a^2 \cot (c+d x) \csc (c+d x)}{3 d (1-\sin (c+d x))}+\frac{a^4 \cot (c+d x) \csc (c+d x)}{3 d (a-a \sin (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 2.11198, size = 190, normalized size = 1.52 \[ \frac{a^2 \left (24 \tan \left (\frac{1}{2} (c+d x)\right )-24 \cot \left (\frac{1}{2} (c+d x)\right )-3 \csc ^2\left (\frac{1}{2} (c+d x)\right )+3 \sec ^2\left (\frac{1}{2} (c+d x)\right )+84 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-84 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{160 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{16 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{8}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(-24*Cot[(c + d*x)/2] - 3*Csc[(c + d*x)/2]^2 - 84*Log[Cos[(c + d*x)/2]] + 84*Log[Sin[(c + d*x)/2]] + 3*Se
c[(c + d*x)/2]^2 + 8/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^3 + (160*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 24*Tan[(c + d*x)/2]))/(24*d)

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Maple [A]  time = 0.135, size = 168, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{7\,{a}^{2}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{7\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{a}^{2}}{3\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{8\,{a}^{2}}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{16\,{a}^{2}\cot \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\,{a}^{2}}{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

1/3/d*a^2/cos(d*x+c)^3+7/2/d*a^2/cos(d*x+c)+7/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))+2/3/d*a^2/sin(d*x+c)/cos(d*x+c
)^3+8/3/d*a^2/sin(d*x+c)/cos(d*x+c)-16/3*a^2*cot(d*x+c)/d+1/3/d*a^2/sin(d*x+c)^2/cos(d*x+c)^3-5/6/d*a^2/sin(d*
x+c)^2/cos(d*x+c)

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Maxima [A]  time = 1.19111, size = 216, normalized size = 1.73 \begin{align*} \frac{8 \,{\left (\tan \left (d x + c\right )^{3} - \frac{3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2}{\left (\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 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 )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(8*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a^2 + a^2*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2
- 2)/(cos(d*x + c)^5 - cos(d*x + c)^3) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)) + 2*a^2*(2*(3*c
os(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)))/d

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Fricas [B]  time = 1.46823, size = 1062, normalized size = 8.5 \begin{align*} -\frac{64 \, a^{2} \cos \left (d x + c\right )^{4} + 86 \, a^{2} \cos \left (d x + c\right )^{3} - 54 \, a^{2} \cos \left (d x + c\right )^{2} - 80 \, a^{2} \cos \left (d x + c\right ) - 4 \, a^{2} + 21 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 3 \, 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 )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - 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 ) - 21 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - a^{2} \cos \left (d x + c\right )^{3} - 3 \, 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 )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - 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 (32 \, a^{2} \cos \left (d x + c\right )^{3} - 11 \, a^{2} \cos \left (d x + c\right )^{2} - 38 \, a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{3} - 3 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - 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)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/12*(64*a^2*cos(d*x + c)^4 + 86*a^2*cos(d*x + c)^3 - 54*a^2*cos(d*x + c)^2 - 80*a^2*cos(d*x + c) - 4*a^2 + 2
1*(a^2*cos(d*x + c)^4 - a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) + 2*a^2 + (a^2*cos(d*x +
c)^3 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 21*(a^2*co
s(d*x + c)^4 - a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) + 2*a^2 + (a^2*cos(d*x + c)^3 + 2*
a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(32*a^2*cos(d*x
+ c)^3 - 11*a^2*cos(d*x + c)^2 - 38*a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 - d*cos(d*x + c)
^3 - 3*d*cos(d*x + c)^2 + d*cos(d*x + c) + (d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - 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)**3*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.34292, size = 203, normalized size = 1.62 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 84 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{3 \,{\left (42 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} - \frac{16 \,{\left (12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 21 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*a^2*tan(1/2*d*x + 1/2*c)^2 + 84*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 24*a^2*tan(1/2*d*x + 1/2*c) - 3*(
42*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^2 - 16*(12*a^2*tan(1/2*
d*x + 1/2*c)^2 - 21*a^2*tan(1/2*d*x + 1/2*c) + 11*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d