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

Optimal. Leaf size=61 \[ -\frac {a A \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {a A \cot (c+d x) \csc (c+d x)}{8 d} \]

[Out]

-1/8*a*A*arctanh(cos(d*x+c))/d-1/8*a*A*cot(d*x+c)*csc(d*x+c)/d+1/4*a*A*cot(d*x+c)*csc(d*x+c)^3/d

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Rubi [A]  time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3962, 2611, 3768, 3770} \[ -\frac {a A \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {a A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {a A \cot (c+d x) \csc (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*A*ArcTanh[Cos[c + d*x]])/(8*d) - (a*A*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (a*A*Cot[c + d*x]*Csc[c + d*x]^3)
/(4*d)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*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 3962

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_))^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[(g*csc[e + f*x])^p*cot[e + f*x]^(2*m), (c
+ d*csc[e + f*x])^(n - m), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 117, normalized size = 1.92 \[ -a A \left (-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*A*(Csc[(c + d*x)/2]^2/(32*d) - Csc[(c + d*x)/2]^4/(64*d) + Log[Cos[(c + d*x)/2]]/(8*d) - Log[Sin[(c + d*x)
/2]]/(8*d) - Sec[(c + d*x)/2]^2/(32*d) + Sec[(c + d*x)/2]^4/(64*d)))

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fricas [B]  time = 0.49, size = 129, normalized size = 2.11 \[ \frac {2 \, A a \cos \left (d x + c\right )^{3} + 2 \, A a \cos \left (d x + c\right ) - {\left (A a \cos \left (d x + c\right )^{4} - 2 \, A a \cos \left (d x + c\right )^{2} + A a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (A a \cos \left (d x + c\right )^{4} - 2 \, A a \cos \left (d x + c\right )^{2} + A a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*A*a*cos(d*x + c)^3 + 2*A*a*cos(d*x + c) - (A*a*cos(d*x + c)^4 - 2*A*a*cos(d*x + c)^2 + A*a)*log(1/2*co
s(d*x + c) + 1/2) + (A*a*cos(d*x + c)^4 - 2*A*a*cos(d*x + c)^2 + A*a)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x
+ c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.40, size = 107, normalized size = 1.75 \[ \frac {4 \, A a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac {A a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (A a - \frac {2 \, A a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*A*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - A*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2
+ (A*a - 2*A*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2)/d

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maple [A]  time = 1.02, size = 65, normalized size = 1.07 \[ \frac {a A \cot \left (d x +c \right ) \left (\csc ^{3}\left (d x +c \right )\right )}{4 d}-\frac {a A \cot \left (d x +c \right ) \csc \left (d x +c \right )}{8 d}+\frac {a A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*A*cot(d*x+c)*csc(d*x+c)^3/d-1/8*a*A*cot(d*x+c)*csc(d*x+c)/d+1/8/d*a*A*ln(csc(d*x+c)-cot(d*x+c))

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maxima [B]  time = 0.32, size = 120, normalized size = 1.97 \[ -\frac {A a {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 4 \, A a {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(A*a*(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)) - 4*A*a*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(co
s(d*x + c) - 1)))/d

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mupad [B]  time = 0.24, size = 56, normalized size = 0.92 \[ \frac {A\,a\,\left (8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+1\right )}{64\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a*(8*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^8 + 1))/(64*d*tan(c/2 + (d*x)/2)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - A a \left (\int \left (- \csc ^{3}{\left (c + d x \right )}\right )\, dx + \int \csc ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*(a+a*csc(d*x+c))*(A-A*csc(d*x+c)),x)

[Out]

-A*a*(Integral(-csc(c + d*x)**3, x) + Integral(csc(c + d*x)**5, x))

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