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

Optimal. Leaf size=128 \[ -\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{6 a^3 \cot (c+d x)}{d}+\frac{23 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{17 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

(-17*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (6*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (3*a^3*Cot[c + d
*x]*Csc[c + d*x])/(2*d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) + (23*a^3*Cos[c + d*x])/(3*d*(1 - Si
n[c + d*x]))

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Rubi [A]  time = 0.206043, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 3768, 2650, 2648} \[ -\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{6 a^3 \cot (c+d x)}{d}+\frac{23 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{17 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-17*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (6*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (3*a^3*Cot[c + d
*x]*Csc[c + d*x])/(2*d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])^2) + (23*a^3*Cos[c + d*x])/(3*d*(1 - Si
n[c + d*x]))

Rule 2872

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

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]

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 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \csc ^4(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^4 \int \left (\frac{7 \csc (c+d x)}{a}+\frac{5 \csc ^2(c+d x)}{a}+\frac{3 \csc ^3(c+d x)}{a}+\frac{\csc ^4(c+d x)}{a}+\frac{2}{a (-1+\sin (c+d x))^2}-\frac{7}{a (-1+\sin (c+d x))}\right ) \, dx\\ &=a^3 \int \csc ^4(c+d x) \, dx+\left (2 a^3\right ) \int \frac{1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \csc ^3(c+d x) \, dx+\left (5 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (7 a^3\right ) \int \csc (c+d x) \, dx-\left (7 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx\\ &=-\frac{7 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac{7 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx+\frac{1}{2} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (5 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac{17 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{6 a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac{23 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.18308, size = 287, normalized size = 2.24 \[ a^3 \left (\frac{17 \tan \left (\frac{1}{2} (c+d x)\right )}{6 d}-\frac{17 \cot \left (\frac{1}{2} (c+d x)\right )}{6 d}-\frac{3 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{3 \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{17 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{17 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{46 \sin \left (\frac{1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 \sin \left (\frac{1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a^3*((-17*Cot[(c + d*x)/2])/(6*d) - (3*Csc[(c + d*x)/2]^2)/(8*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d
) - (17*Log[Cos[(c + d*x)/2]])/(2*d) + (17*Log[Sin[(c + d*x)/2]])/(2*d) + (3*Sec[(c + d*x)/2]^2)/(8*d) + 2/(3*
d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*Sin[(c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
) + (46*Sin[(c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (17*Tan[(c + d*x)/2])/(6*d) + (Sec[(c
+ d*x)/2]^2*Tan[(c + d*x)/2])/(24*d))

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Maple [A]  time = 0.164, size = 214, normalized size = 1.7 \begin{align*}{\frac{{a}^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{17\,{a}^{3}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{17\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{3}}{d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{20\,{a}^{3}}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{40\,{a}^{3}\cot \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\,{a}^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{{a}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{2\,{a}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*a^3/cos(d*x+c)^3+17/2/d*a^3/cos(d*x+c)+17/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))+1/d*a^3/sin(d*x+c)/cos(d*x+c
)^3+20/3/d*a^3/sin(d*x+c)/cos(d*x+c)-40/3*a^3*cot(d*x+c)/d+1/d*a^3/sin(d*x+c)^2/cos(d*x+c)^3-5/2/d*a^3/sin(d*x
+c)^2/cos(d*x+c)+1/3/d*a^3/sin(d*x+c)^3/cos(d*x+c)^3-2/3/d*a^3/sin(d*x+c)^3/cos(d*x+c)

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Maxima [A]  time = 1.20841, size = 277, normalized size = 2.16 \begin{align*} \frac{12 \,{\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^{3} + 4 \,{\left (\tan \left (d x + c\right )^{3} - \frac{9 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} + 9 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, a^{3}{\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^{3}{\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)^4*sec(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(12*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a^3 + 4*(tan(d*x + c)^3 - (9*tan(d*x + c)^2 + 1)/t
an(d*x + c)^3 + 9*tan(d*x + c))*a^3 + 3*a^3*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2)/(cos(d*x + c)^5 - c
os(d*x + c)^3) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)) + 2*a^3*(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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Fricas [B]  time = 1.52212, size = 1314, normalized size = 10.27 \begin{align*} \frac{160 \, a^{3} \cos \left (d x + c\right )^{5} - 58 \, a^{3} \cos \left (d x + c\right )^{4} - 356 \, a^{3} \cos \left (d x + c\right )^{3} + 70 \, a^{3} \cos \left (d x + c\right )^{2} + 200 \, a^{3} \cos \left (d x + c\right ) - 8 \, a^{3} - 51 \,{\left (a^{3} \cos \left (d x + c\right )^{5} + 2 \, a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} - 4 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{4} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 51 \,{\left (a^{3} \cos \left (d x + c\right )^{5} + 2 \, a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} - 4 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{4} - a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\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 (80 \, a^{3} \cos \left (d x + c\right )^{4} + 109 \, a^{3} \cos \left (d x + c\right )^{3} - 69 \, a^{3} \cos \left (d x + c\right )^{2} - 104 \, a^{3} \cos \left (d x + c\right ) - 4 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{5} + 2 \, d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) -{\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 ) + 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)^4*sec(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(160*a^3*cos(d*x + c)^5 - 58*a^3*cos(d*x + c)^4 - 356*a^3*cos(d*x + c)^3 + 70*a^3*cos(d*x + c)^2 + 200*a^
3*cos(d*x + c) - 8*a^3 - 51*(a^3*cos(d*x + c)^5 + 2*a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^3 - 4*a^3*cos(d*x
+ c)^2 + a^3*cos(d*x + c) + 2*a^3 - (a^3*cos(d*x + c)^4 - a^3*cos(d*x + c)^3 - 3*a^3*cos(d*x + c)^2 + a^3*cos(
d*x + c) + 2*a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 51*(a^3*cos(d*x + c)^5 + 2*a^3*cos(d*x + c)^4 -
2*a^3*cos(d*x + c)^3 - 4*a^3*cos(d*x + c)^2 + a^3*cos(d*x + c) + 2*a^3 - (a^3*cos(d*x + c)^4 - a^3*cos(d*x + c
)^3 - 3*a^3*cos(d*x + c)^2 + a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(80*a^3*
cos(d*x + c)^4 + 109*a^3*cos(d*x + c)^3 - 69*a^3*cos(d*x + c)^2 - 104*a^3*cos(d*x + c) - 4*a^3)*sin(d*x + c))/
(d*cos(d*x + c)^5 + 2*d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3 - 4*d*cos(d*x + c)^2 + d*cos(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) + 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)**4*sec(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21421, size = 262, normalized size = 2.05 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 204 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 69 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{187 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 60 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 405 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 394 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c)^2 + 204*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 69*
a^3*tan(1/2*d*x + 1/2*c) - (187*a^3*tan(1/2*d*x + 1/2*c)^6 - 60*a^3*tan(1/2*d*x + 1/2*c)^5 - 405*a^3*tan(1/2*d
*x + 1/2*c)^4 + 394*a^3*tan(1/2*d*x + 1/2*c)^3 - 45*a^3*tan(1/2*d*x + 1/2*c)^2 - 6*a^3*tan(1/2*d*x + 1/2*c) -
a^3)/(tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c))^3)/d