3.233 \(\int \csc ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac{a^3 A \cot ^5(c+d x)}{5 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac{a^3 A \cot (c+d x) \csc (c+d x)}{4 d} \]

[Out]

(a^3*A*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (a^3*A*Cot[c + d*x]^5)/(5*d) + (a^3*A*C
ot[c + d*x]*Csc[c + d*x])/(4*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d)

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Rubi [A]  time = 0.234037, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2950, 2709, 3767, 8, 3768, 3770} \[ -\frac{a^3 A \cot ^5(c+d x)}{5 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac{a^3 A \cot (c+d x) \csc (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*ArcTanh[Cos[c + d*x]])/(4*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (a^3*A*Cot[c + d*x]^5)/(5*d) + (a^3*A*C
ot[c + d*x]*Csc[c + d*x])/(4*d) - (a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d)

Rule 2950

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

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 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 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 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 ^6(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\left (a^3 A^3\right ) \int \frac{\cot ^6(c+d x)}{(A-A \sin (c+d x))^2} \, dx\\ &=\frac{a^3 \int \left (-A^4 \csc ^2(c+d x)-2 A^4 \csc ^3(c+d x)+2 A^4 \csc ^5(c+d x)+A^4 \csc ^6(c+d x)\right ) \, dx}{A^3}\\ &=-\left (\left (a^3 A\right ) \int \csc ^2(c+d x) \, dx\right )+\left (a^3 A\right ) \int \csc ^6(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^5(c+d x) \, dx\\ &=\frac{a^3 A \cot (c+d x) \csc (c+d x)}{d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}-\left (a^3 A\right ) \int \csc (c+d x) \, dx+\frac{1}{2} \left (3 a^3 A\right ) \int \csc ^3(c+d x) \, dx+\frac{\left (a^3 A\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{a^3 A \cot ^5(c+d x)}{5 d}+\frac{a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac{1}{4} \left (3 a^3 A\right ) \int \csc (c+d x) \, dx\\ &=\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{a^3 A \cot ^5(c+d x)}{5 d}+\frac{a^3 A \cot (c+d x) \csc (c+d x)}{4 d}-\frac{a^3 A \cot (c+d x) \csc ^3(c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.074181, size = 268, normalized size = 2.55 \[ a^3 A \left (-\frac{7 \tan \left (\frac{1}{2} (c+d x)\right )}{30 d}+\frac{7 \cot \left (\frac{1}{2} (c+d x)\right )}{30 d}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{16 d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{32 d}-\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{16 d}-\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}+\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{160 d}-\frac{19 \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{480 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )}{160 d}+\frac{19 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{480 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*((7*Cot[(c + d*x)/2])/(30*d) + Csc[(c + d*x)/2]^2/(16*d) - (19*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(480
*d) - Csc[(c + d*x)/2]^4/(32*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(160*d) + Log[Cos[(c + d*x)/2]]/(4*d)
- Log[Sin[(c + d*x)/2]]/(4*d) - Sec[(c + d*x)/2]^2/(16*d) + Sec[(c + d*x)/2]^4/(32*d) - (7*Tan[(c + d*x)/2])/(
30*d) + (19*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(480*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(160*d))

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Maple [A]  time = 0.06, size = 132, normalized size = 1.3 \begin{align*}{\frac{7\,{a}^{3}A\cot \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{3}A\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{4\,d}}-{\frac{{a}^{3}A\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

7/15*a^3*A*cot(d*x+c)/d+1/4*a^3*A*cot(d*x+c)*csc(d*x+c)/d-1/4/d*a^3*A*ln(csc(d*x+c)-cot(d*x+c))-1/2*a^3*A*cot(
d*x+c)*csc(d*x+c)^3/d-1/5/d*a^3*A*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^3*A*cot(d*x+c)*csc(d*x+c)^2

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Maxima [A]  time = 0.994365, size = 236, normalized size = 2.25 \begin{align*} \frac{15 \, A a^{3}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{120 \, A a^{3}}{\tan \left (d x + c\right )} - \frac{8 \,{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} A a^{3}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/120*(15*A*a^3*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x
+ c) + 1) + 3*log(cos(d*x + c) - 1)) - 60*A*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1)
+ log(cos(d*x + c) - 1)) + 120*A*a^3/tan(d*x + c) - 8*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*A*a^3/tan(d*
x + c)^5)/d

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Fricas [B]  time = 2.09352, size = 520, normalized size = 4.95 \begin{align*} \frac{56 \, A a^{3} \cos \left (d x + c\right )^{5} - 80 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \,{\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 15 \,{\left (A a^{3} \cos \left (d x + c\right )^{4} - 2 \, A a^{3} \cos \left (d x + c\right )^{2} + A a^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left (A a^{3} \cos \left (d x + c\right )^{3} + A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(56*A*a^3*cos(d*x + c)^5 - 80*A*a^3*cos(d*x + c)^3 + 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 +
A*a^3)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 15*(A*a^3*cos(d*x + c)^4 - 2*A*a^3*cos(d*x + c)^2 + A*a^3)*
log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 30*(A*a^3*cos(d*x + c)^3 + A*a^3*cos(d*x + c))*sin(d*x + c))/((d*c
os(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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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)**6*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.20684, size = 235, normalized size = 2.24 \begin{align*} \frac{3 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 90 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{274 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 90 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 25 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/480*(3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 12
0*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 90*A*a^3*tan(1/2*d*x + 1/2*c) + (274*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 9
0*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 25*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 3*A*a^3)/ta
n(1/2*d*x + 1/2*c)^5)/d