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

Optimal. Leaf size=110 \[ -\frac{3 a^3 \cot (c+d x)}{d}+\frac{17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

(-11*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a^3*Cot[c + d*x])/d - (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) + (17*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x]))

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Rubi [A]  time = 0.190362, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{3 a^3 \cot (c+d x)}{d}+\frac{17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a^3*Cot[c + d*x])/d - (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) + (17*a^3*Cos[c + d*x])/(3*d*(1 - Sin[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 ^3(c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^4 \int \left (\frac{5 \csc (c+d x)}{a}+\frac{3 \csc ^2(c+d x)}{a}+\frac{\csc ^3(c+d x)}{a}+\frac{2}{a (-1+\sin (c+d x))^2}-\frac{5}{a (-1+\sin (c+d x))}\right ) \, dx\\ &=a^3 \int \csc ^3(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 ^2(c+d x) \, dx+\left (5 a^3\right ) \int \csc (c+d x) \, dx-\left (5 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx\\ &=-\frac{5 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{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{5 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{1}{2} a^3 \int \csc (c+d x) \, dx-\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac{11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x)}{d}-\frac{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{17 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.11083, size = 190, normalized size = 1.73 \[ \frac{a^3 \left (36 \tan \left (\frac{1}{2} (c+d x)\right )-36 \cot \left (\frac{1}{2} (c+d x)\right )-3 \csc ^2\left (\frac{1}{2} (c+d x)\right )+3 \sec ^2\left (\frac{1}{2} (c+d x)\right )+132 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-132 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{272 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{32 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{16}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-36*Cot[(c + d*x)/2] - 3*Csc[(c + d*x)/2]^2 - 132*Log[Cos[(c + d*x)/2]] + 132*Log[Sin[(c + d*x)/2]] + 3*
Sec[(c + d*x)/2]^2 + 16/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (32*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2])^3 + (272*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 36*Tan[(c + d*x)/2]))/(24*d
)

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Maple [A]  time = 0.244, size = 202, normalized size = 1.8 \begin{align*}{\frac{2\,{a}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{11\,{a}^{3}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{11\,{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}}}+4\,{\frac{{a}^{3}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-8\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\,{a}^{3}}{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))^3,x)

[Out]

2/3*a^3*tan(d*x+c)/d+1/3/d*a^3*tan(d*x+c)*sec(d*x+c)^2+1/d*a^3/cos(d*x+c)^3+11/2/d*a^3/cos(d*x+c)+11/2/d*a^3*l
n(csc(d*x+c)-cot(d*x+c))+1/d*a^3/sin(d*x+c)/cos(d*x+c)^3+4/d*a^3/sin(d*x+c)/cos(d*x+c)-8*a^3*cot(d*x+c)/d+1/3/
d*a^3/sin(d*x+c)^2/cos(d*x+c)^3-5/6/d*a^3/sin(d*x+c)^2/cos(d*x+c)

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Maxima [A]  time = 1.18138, size = 246, normalized size = 2.24 \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} + 3 \, \tan \left (d x + c\right )\right )} a^{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 )} + 6 \, 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)^3*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 + 3*tan(d*x + c))*a^3 + a^
3*(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)) + 6*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 = 1.4775, size = 1067, normalized size = 9.7 \begin{align*} -\frac{104 \, a^{3} \cos \left (d x + c\right )^{4} + 142 \, a^{3} \cos \left (d x + c\right )^{3} - 90 \, a^{3} \cos \left (d x + c\right )^{2} - 136 \, a^{3} \cos \left (d x + c\right ) - 8 \, a^{3} + 33 \,{\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} +{\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, 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 ) - 33 \,{\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} +{\left (a^{3} \cos \left (d x + c\right )^{3} + 2 \, 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 (52 \, a^{3} \cos \left (d x + c\right )^{3} - 19 \, a^{3} \cos \left (d x + c\right )^{2} - 64 \, 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 )^{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))^3,x, algorithm="fricas")

[Out]

-1/12*(104*a^3*cos(d*x + c)^4 + 142*a^3*cos(d*x + c)^3 - 90*a^3*cos(d*x + c)^2 - 136*a^3*cos(d*x + c) - 8*a^3
+ 33*(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 + (a^3*cos(d*x
+ c)^3 + 2*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) - 33*(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 + (a^3*cos(d*x + c)^3 +
2*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*(52*a^3*cos(d
*x + c)^3 - 19*a^3*cos(d*x + c)^2 - 64*a^3*cos(d*x + c) + 4*a^3)*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)*s
in(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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20275, size = 203, normalized size = 1.85 \begin{align*} \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 132 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{3 \,{\left (66 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} - \frac{16 \,{\left (21 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 19 \, a^{3}\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))^3,x, algorithm="giac")

[Out]

1/24*(3*a^3*tan(1/2*d*x + 1/2*c)^2 + 132*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 36*a^3*tan(1/2*d*x + 1/2*c) - 3*
(66*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^2 - 16*(21*a^3*tan(1/
2*d*x + 1/2*c)^2 - 36*a^3*tan(1/2*d*x + 1/2*c) + 19*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d