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

Optimal. Leaf size=87 \[ \frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {10 a^2 \cot (c+d x)}{3 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))} \]

[Out]

-2*a^2*arctanh(cos(d*x+c))/d-10/3*a^2*cot(d*x+c)/d+2*a^2*cot(d*x+c)/d/(1-sin(d*x+c))+1/3*a^4*cot(d*x+c)/d/(a-a
*sin(d*x+c))^2

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Rubi [A]  time = 0.27, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2869, 2766, 2978, 2748, 3767, 8, 3770} \[ -\frac {10 a^2 \cot (c+d x)}{3 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*ArcTanh[Cos[c + d*x]])/d - (10*a^2*Cot[c + d*x])/(3*d) + (2*a^2*Cot[c + d*x])/(d*(1 - Sin[c + d*x])) +
(a^4*Cot[c + d*x])/(3*d*(a - a*Sin[c + d*x])^2)

Rule 8

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

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 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 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 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 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 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 ^2(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx &=a^4 \int \frac {\csc ^2(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\csc ^2(c+d x) (4 a+2 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} \int \csc ^2(c+d x) \left (10 a^2+6 a^2 \sin (c+d x)\right ) \, dx\\ &=\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\frac {1}{3} \left (10 a^2\right ) \int \csc ^2(c+d x) \, dx\\ &=-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\left (10 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d}\\ &=-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {10 a^2 \cot (c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d (1-\sin (c+d x))}+\frac {a^4 \cot (c+d x)}{3 d (a-a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.92, size = 135, normalized size = 1.55 \[ \frac {a^2 \left (3 \tan \left (\frac {1}{2} (c+d x)\right )-3 \cot \left (\frac {1}{2} (c+d x)\right )+12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right ) (7 \sin (c+d x)-8)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-3*Cot[(c + d*x)/2] - 12*Log[Cos[(c + d*x)/2]] + 12*Log[Sin[(c + d*x)/2]] + 2/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2 + (4*Sin[(c + d*x)/2]*(-8 + 7*Sin[c + d*x]))/(-Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 3*Tan[(c
+ d*x)/2]))/(6*d)

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fricas [B]  time = 0.48, size = 329, normalized size = 3.78 \[ \frac {10 \, a^{2} \cos \left (d x + c\right )^{3} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 13 \, a^{2} \cos \left (d x + c\right ) + a^{2} - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (10 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(10*a^2*cos(d*x + c)^3 - 4*a^2*cos(d*x + c)^2 - 13*a^2*cos(d*x + c) + a^2 - 3*(a^2*cos(d*x + c)^3 + 2*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*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)^3 + 2*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*cos(d*x + c) - 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + (10*a^2*cos(d*x + c)^2 + 14
*a^2*cos(d*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c) - (d*cos(d*x +
c)^2 - d*cos(d*x + c) - 2*d)*sin(d*x + c) - 2*d)

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giac [A]  time = 0.22, size = 118, normalized size = 1.36 \[ \frac {12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^2*tan(1/2*d*x + 1/2*c) - 3*(4*a^2*tan(1/2*d*x + 1/2*c) + a^2)
/tan(1/2*d*x + 1/2*c) - 4*(9*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) + 8*a^2)/(tan(1/2*d*x +
1/2*c) - 1)^3)/d

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maple [A]  time = 0.73, size = 156, normalized size = 1.79 \[ \frac {2 a^{2} \tan \left (d x +c \right )}{3 d}+\frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2}}{3 d \cos \left (d x +c \right )^{3}}+\frac {2 a^{2}}{d \cos \left (d x +c \right )}+\frac {2 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a^{2}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}+\frac {4 a^{2}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 a^{2} \cot \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 107, normalized size = 1.23 \[ \frac {{\left (\tan \left (d x + c\right )^{3} - \frac {3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a^{2} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.45, size = 144, normalized size = 1.66 \[ \frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+23\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {41\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a^2}{d\,\left (-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^4*sin(c + d*x)^2),x)

[Out]

(2*a^2*log(tan(c/2 + (d*x)/2)))/d - (23*a^2*tan(c/2 + (d*x)/2)^2 - 13*a^2*tan(c/2 + (d*x)/2)^3 + a^2 - (41*a^2
*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c/2 + (d*x)/2) - 6*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2
+ (d*x)/2)^4)) + (a^2*tan(c/2 + (d*x)/2))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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