3.234 \(\int \csc ^7(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\)

Optimal. Leaf size=130 \[ -\frac{2 a^3 A \cot ^5(c+d x)}{5 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{3 a^3 A \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac{5 a^3 A \cot (c+d x) \csc ^3(c+d x)}{24 d}+\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{16 d} \]

[Out]

(3*a^3*A*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (2*a^3*A*Cot[c + d*x]^5)/(5*d) + (3*
a^3*A*Cot[c + d*x]*Csc[c + d*x])/(16*d) - (5*a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(24*d) - (a^3*A*Cot[c + d*x]*C
sc[c + d*x]^5)/(6*d)

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Rubi [A]  time = 0.19665, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 3768, 3770, 3767} \[ -\frac{2 a^3 A \cot ^5(c+d x)}{5 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}+\frac{3 a^3 A \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac{5 a^3 A \cot (c+d x) \csc ^3(c+d x)}{24 d}+\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(3*a^3*A*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a^3*A*Cot[c + d*x]^3)/(3*d) - (2*a^3*A*Cot[c + d*x]^5)/(5*d) + (3*
a^3*A*Cot[c + d*x]*Csc[c + d*x])/(16*d) - (5*a^3*A*Cot[c + d*x]*Csc[c + d*x]^3)/(24*d) - (a^3*A*Cot[c + d*x]*C
sc[c + d*x]^5)/(6*d)

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

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]

Rubi steps

\begin{align*} \int \csc ^7(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\int \left (-a^3 A \csc ^3(c+d x)-2 a^3 A \csc ^4(c+d x)+2 a^3 A \csc ^6(c+d x)+a^3 A \csc ^7(c+d x)\right ) \, dx\\ &=-\left (\left (a^3 A\right ) \int \csc ^3(c+d x) \, dx\right )+\left (a^3 A\right ) \int \csc ^7(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc ^4(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^6(c+d x) \, dx\\ &=\frac{a^3 A \cot (c+d x) \csc (c+d x)}{2 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}-\frac{1}{2} \left (a^3 A\right ) \int \csc (c+d x) \, dx+\frac{1}{6} \left (5 a^3 A\right ) \int \csc ^5(c+d x) \, dx+\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (2 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))}{2 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{2 a^3 A \cot ^5(c+d x)}{5 d}+\frac{a^3 A \cot (c+d x) \csc (c+d x)}{2 d}-\frac{5 a^3 A \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}+\frac{1}{8} \left (5 a^3 A\right ) \int \csc ^3(c+d x) \, dx\\ &=\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{2 a^3 A \cot ^5(c+d x)}{5 d}+\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{16 d}-\frac{5 a^3 A \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}+\frac{1}{16} \left (5 a^3 A\right ) \int \csc (c+d x) \, dx\\ &=\frac{3 a^3 A \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac{2 a^3 A \cot ^3(c+d x)}{3 d}-\frac{2 a^3 A \cot ^5(c+d x)}{5 d}+\frac{3 a^3 A \cot (c+d x) \csc (c+d x)}{16 d}-\frac{5 a^3 A \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac{a^3 A \cot (c+d x) \csc ^5(c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 0.0797904, size = 306, normalized size = 2.35 \[ a^3 A \left (-\frac{2 \tan \left (\frac{1}{2} (c+d x)\right )}{15 d}+\frac{2 \cot \left (\frac{1}{2} (c+d x)\right )}{15 d}-\frac{\csc ^6\left (\frac{1}{2} (c+d x)\right )}{384 d}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{3 \csc ^2\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{\sec ^6\left (\frac{1}{2} (c+d x)\right )}{384 d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{3 \sec ^2\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{16 d}+\frac{3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{16 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{80 d}+\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{240 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )}{80 d}-\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{240 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*((2*Cot[(c + d*x)/2])/(15*d) + (3*Csc[(c + d*x)/2]^2)/(64*d) + (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24
0*d) - Csc[(c + d*x)/2]^4/(64*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(80*d) - Csc[(c + d*x)/2]^6/(384*d) +
(3*Log[Cos[(c + d*x)/2]])/(16*d) - (3*Log[Sin[(c + d*x)/2]])/(16*d) - (3*Sec[(c + d*x)/2]^2)/(64*d) + Sec[(c
+ d*x)/2]^4/(64*d) + Sec[(c + d*x)/2]^6/(384*d) - (2*Tan[(c + d*x)/2])/(15*d) - (Sec[(c + d*x)/2]^2*Tan[(c + d
*x)/2])/(240*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(80*d))

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Maple [A]  time = 0.062, size = 155, normalized size = 1.2 \begin{align*}{\frac{3\,{a}^{3}A\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{16\,d}}-{\frac{3\,{a}^{3}A\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{4\,{a}^{3}A\cot \left ( dx+c \right ) }{15\,d}}+{\frac{2\,{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{2\,{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{5}}{6\,d}}-{\frac{5\,{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{24\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^7*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x)

[Out]

3/16*a^3*A*cot(d*x+c)*csc(d*x+c)/d-3/16/d*a^3*A*ln(csc(d*x+c)-cot(d*x+c))+4/15*a^3*A*cot(d*x+c)/d+2/15/d*a^3*A
*cot(d*x+c)*csc(d*x+c)^2-2/5/d*a^3*A*cot(d*x+c)*csc(d*x+c)^4-1/6*a^3*A*cot(d*x+c)*csc(d*x+c)^5/d-5/24*a^3*A*co
t(d*x+c)*csc(d*x+c)^3/d

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Maxima [A]  time = 0.990973, size = 279, normalized size = 2.15 \begin{align*} \frac{5 \, A a^{3}{\left (\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 120 \, 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{320 \,{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} A a^{3}}{\tan \left (d x + c\right )^{3}} - \frac{64 \,{\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}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(5*A*a^3*(2*(15*cos(d*x + c)^5 - 40*cos(d*x + c)^3 + 33*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4
+ 3*cos(d*x + c)^2 - 1) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1)) - 120*A*a^3*(2*cos(d*x + c)/(c
os(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 320*(3*tan(d*x + c)^2 + 1)*A*a^3/tan(d*x
+ c)^3 - 64*(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 = 1.95144, size = 602, normalized size = 4.63 \begin{align*} -\frac{90 \, A a^{3} \cos \left (d x + c\right )^{5} - 80 \, A a^{3} \cos \left (d x + c\right )^{3} - 90 \, A a^{3} \cos \left (d x + c\right ) - 45 \,{\left (A a^{3} \cos \left (d x + c\right )^{6} - 3 \, A a^{3} \cos \left (d x + c\right )^{4} + 3 \, 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 ) + 45 \,{\left (A a^{3} \cos \left (d x + c\right )^{6} - 3 \, A a^{3} \cos \left (d x + c\right )^{4} + 3 \, 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 ) + 64 \,{\left (2 \, A a^{3} \cos \left (d x + c\right )^{5} - 5 \, A a^{3} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{480 \,{\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \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)^7*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/480*(90*A*a^3*cos(d*x + c)^5 - 80*A*a^3*cos(d*x + c)^3 - 90*A*a^3*cos(d*x + c) - 45*(A*a^3*cos(d*x + c)^6 -
3*A*a^3*cos(d*x + c)^4 + 3*A*a^3*cos(d*x + c)^2 - A*a^3)*log(1/2*cos(d*x + c) + 1/2) + 45*(A*a^3*cos(d*x + c)
^6 - 3*A*a^3*cos(d*x + c)^4 + 3*A*a^3*cos(d*x + c)^2 - A*a^3)*log(-1/2*cos(d*x + c) + 1/2) + 64*(2*A*a^3*cos(d
*x + c)^5 - 5*A*a^3*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(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)**7*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.21561, size = 327, normalized size = 2.52 \begin{align*} \frac{5 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 24 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 45 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 40 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 360 \, A a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 240 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{882 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 240 \, 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} - 40 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 45 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, A a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1920*(5*A*a^3*tan(1/2*d*x + 1/2*c)^6 + 24*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 45*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 4
0*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 360*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) -
240*A*a^3*tan(1/2*d*x + 1/2*c) + (882*A*a^3*tan(1/2*d*x + 1/2*c)^6 + 240*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 - 40*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 45*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*A*a^3*tan(1
/2*d*x + 1/2*c) - 5*A*a^3)/tan(1/2*d*x + 1/2*c)^6)/d