3.35 \(\int (c+d x) \csc ^3(a+b x) \, dx\)

Optimal. Leaf size=109 \[ \frac {i d \text {Li}_2\left (-e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (e^{i (a+b x)}\right )}{2 b^2}-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b} \]

[Out]

-(d*x+c)*arctanh(exp(I*(b*x+a)))/b-1/2*d*csc(b*x+a)/b^2-1/2*(d*x+c)*cot(b*x+a)*csc(b*x+a)/b+1/2*I*d*polylog(2,
-exp(I*(b*x+a)))/b^2-1/2*I*d*polylog(2,exp(I*(b*x+a)))/b^2

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Rubi [A]  time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4185, 4183, 2279, 2391} \[ \frac {i d \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Csc[a + b*x]^3,x]

[Out]

-(((c + d*x)*ArcTanh[E^(I*(a + b*x))])/b) - (d*Csc[a + b*x])/(2*b^2) - ((c + d*x)*Cot[a + b*x]*Csc[a + b*x])/(
2*b) + ((I/2)*d*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((I/2)*d*PolyLog[2, E^(I*(a + b*x))])/b^2

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

\begin {align*} \int (c+d x) \csc ^3(a+b x) \, dx &=-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b}+\frac {1}{2} \int (c+d x) \csc (a+b x) \, dx\\ &=-\frac {(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b}-\frac {d \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {d \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=-\frac {(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=-\frac {(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \csc (a+b x)}{2 b^2}-\frac {(c+d x) \cot (a+b x) \csc (a+b x)}{2 b}+\frac {i d \text {Li}_2\left (-e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (e^{i (a+b x)}\right )}{2 b^2}\\ \end {align*}

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Mathematica [B]  time = 2.03, size = 292, normalized size = 2.68 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i (a+b x)}\right )-\text {Li}_2\left (e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )-a \log \left (\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )}{2 b^2}+\frac {d \csc \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right ) \csc \left (\frac {a}{2}+\frac {b x}{2}\right )}{4 b^2}-\frac {d \sec \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right )}{4 b^2}-\frac {c \csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {c \sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {c \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {c \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {d x \csc ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {d x \sec ^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Csc[a + b*x]^3,x]

[Out]

-1/8*(d*x*Csc[a/2 + (b*x)/2]^2)/b - (c*Csc[(a + b*x)/2]^2)/(8*b) - (c*Log[Cos[(a + b*x)/2]])/(2*b) + (c*Log[Si
n[(a + b*x)/2]])/(2*b) + (d*((a + b*x)*(Log[1 - E^(I*(a + b*x))] - Log[1 + E^(I*(a + b*x))]) - a*Log[Tan[(a +
b*x)/2]] + I*(PolyLog[2, -E^(I*(a + b*x))] - PolyLog[2, E^(I*(a + b*x))])))/(2*b^2) + (d*x*Sec[a/2 + (b*x)/2]^
2)/(8*b) + (c*Sec[(a + b*x)/2]^2)/(8*b) + (d*Csc[a/2]*Csc[a/2 + (b*x)/2]*Sin[(b*x)/2])/(4*b^2) - (d*Sec[a/2]*S
ec[a/2 + (b*x)/2]*Sin[(b*x)/2])/(4*b^2)

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fricas [B]  time = 0.56, size = 452, normalized size = 4.15 \[ \frac {2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) + {\left (-i \, d \cos \left (b x + a\right )^{2} + i \, d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (i \, d \cos \left (b x + a\right )^{2} - i \, d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (-i \, d \cos \left (b x + a\right )^{2} + i \, d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (i \, d \cos \left (b x + a\right )^{2} - i \, d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (b d x - {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b d x - {\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + {\left ({\left (b c - a d\right )} \cos \left (b x + a\right )^{2} - b c + a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left ({\left (b c - a d\right )} \cos \left (b x + a\right )^{2} - b c + a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - {\left (b d x - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - {\left (b d x - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + 2 \, d \sin \left (b x + a\right )}{4 \, {\left (b^{2} \cos \left (b x + a\right )^{2} - b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(b*d*x + b*c)*cos(b*x + a) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(cos(b*x + a) + I*sin(b*x + a)) + (I*d*co
s(b*x + a)^2 - I*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(-cos(b*x + a) + I
*sin(b*x + a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(-cos(b*x + a) - I*sin(b*x + a)) + (b*d*x - (b*d*x + b*c)*cos
(b*x + a)^2 + b*c)*log(cos(b*x + a) + I*sin(b*x + a) + 1) + (b*d*x - (b*d*x + b*c)*cos(b*x + a)^2 + b*c)*log(c
os(b*x + a) - I*sin(b*x + a) + 1) + ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos(b*x + a) + 1/2*I*sin
(b*x + a) + 1/2) + ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2)
- (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - (b*d*x - (b*d*x + a*d
)*cos(b*x + a)^2 + a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) + 2*d*sin(b*x + a))/(b^2*cos(b*x + a)^2 - b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \csc \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*csc(b*x + a)^3, x)

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maple [B]  time = 0.09, size = 246, normalized size = 2.26 \[ \frac {d x b \,{\mathrm e}^{3 i \left (b x +a \right )}+c b \,{\mathrm e}^{3 i \left (b x +a \right )}+d x b \,{\mathrm e}^{i \left (b x +a \right )}+c b \,{\mathrm e}^{i \left (b x +a \right )}-i d \,{\mathrm e}^{3 i \left (b x +a \right )}+i d \,{\mathrm e}^{i \left (b x +a \right )}}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {c \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{2 b}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{2 b^{2}}+\frac {i d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}-\frac {i d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {d a \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2/(exp(2*I*(b*x+a))-1)^2*(d*x*b*exp(3*I*(b*x+a))+c*b*exp(3*I*(b*x+a))+d*x*b*exp(I*(b*x+a))+c*b*exp(I*(b*x+
a))-I*d*exp(3*I*(b*x+a))+I*d*exp(I*(b*x+a)))-1/b*c*arctanh(exp(I*(b*x+a)))-1/2/b*d*ln(exp(I*(b*x+a))+1)*x-1/2/
b^2*d*ln(exp(I*(b*x+a))+1)*a+1/2*I*d*polylog(2,-exp(I*(b*x+a)))/b^2+1/2/b*d*ln(1-exp(I*(b*x+a)))*x+1/2/b^2*d*l
n(1-exp(I*(b*x+a)))*a-1/2*I*d*polylog(2,exp(I*(b*x+a)))/b^2+1/b^2*d*a*arctanh(exp(I*(b*x+a)))

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maxima [B]  time = 0.54, size = 769, normalized size = 7.06 \[ -\frac {{\left (2 \, b d x + 2 \, b c + 2 \, {\left (b d x + b c\right )} \cos \left (4 \, b x + 4 \, a\right ) - 4 \, {\left (b d x + b c\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (2 i \, b d x + 2 i \, b c\right )} \sin \left (4 \, b x + 4 \, a\right ) + {\left (-4 i \, b d x - 4 i \, b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - {\left (2 \, b c \cos \left (4 \, b x + 4 \, a\right ) - 4 \, b c \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b c \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, b c \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + {\left (2 \, b d x \cos \left (4 \, b x + 4 \, a\right ) - 4 \, b d x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b d x \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, b d x \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b d x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + {\left (4 i \, b d x + 4 i \, b c + 4 \, d\right )} \cos \left (3 \, b x + 3 \, a\right ) + {\left (4 i \, b d x + 4 i \, b c - 4 \, d\right )} \cos \left (b x + a\right ) - {\left (2 \, d \cos \left (4 \, b x + 4 \, a\right ) - 4 \, d \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, d \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, d\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (2 \, d \cos \left (4 \, b x + 4 \, a\right ) - 4 \, d \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, d \sin \left (4 \, b x + 4 \, a\right ) - 4 i \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, d\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left (-i \, b d x - i \, b c + {\left (-i \, b d x - i \, b c\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (2 i \, b d x + 2 i \, b c\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right ) - 2 \, {\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + {\left (i \, b d x + i \, b c + {\left (i \, b d x + i \, b c\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (-2 i \, b d x - 2 i \, b c\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (b d x + b c\right )} \sin \left (4 \, b x + 4 \, a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - 4 \, {\left (b d x + b c - i \, d\right )} \sin \left (3 \, b x + 3 \, a\right ) - 4 \, {\left (b d x + b c + i \, d\right )} \sin \left (b x + a\right )}{-4 i \, b^{2} \cos \left (4 \, b x + 4 \, a\right ) + 8 i \, b^{2} \cos \left (2 \, b x + 2 \, a\right ) + 4 \, b^{2} \sin \left (4 \, b x + 4 \, a\right ) - 8 \, b^{2} \sin \left (2 \, b x + 2 \, a\right ) - 4 i \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(4*b*x + 4*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a) + (2*I*b*d*x + 2*I*b*
c)*sin(4*b*x + 4*a) + (-4*I*b*d*x - 4*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b*
c*cos(4*b*x + 4*a) - 4*b*c*cos(2*b*x + 2*a) + 2*I*b*c*sin(4*b*x + 4*a) - 4*I*b*c*sin(2*b*x + 2*a) + 2*b*c)*arc
tan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(4*b*x + 4*a) - 4*b*d*x*cos(2*b*x + 2*a) + 2*I*b*d*x*sin(4*
b*x + 4*a) - 4*I*b*d*x*sin(2*b*x + 2*a) + 2*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (4*I*b*d*x + 4*I
*b*c + 4*d)*cos(3*b*x + 3*a) + (4*I*b*d*x + 4*I*b*c - 4*d)*cos(b*x + a) - (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b*
x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(-e^(I*b*x + I*a)) + (2*d*cos(4*b*x + 4
*a) - 4*d*cos(2*b*x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(e^(I*b*x + I*a)) + (
-I*b*d*x - I*b*c + (-I*b*d*x - I*b*c)*cos(4*b*x + 4*a) + (2*I*b*d*x + 2*I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*c
)*sin(4*b*x + 4*a) - 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) +
1) + (I*b*d*x + I*b*c + (I*b*d*x + I*b*c)*cos(4*b*x + 4*a) + (-2*I*b*d*x - 2*I*b*c)*cos(2*b*x + 2*a) - (b*d*x
+ b*c)*sin(4*b*x + 4*a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x +
a) + 1) - 4*(b*d*x + b*c - I*d)*sin(3*b*x + 3*a) - 4*(b*d*x + b*c + I*d)*sin(b*x + a))/(-4*I*b^2*cos(4*b*x + 4
*a) + 8*I*b^2*cos(2*b*x + 2*a) + 4*b^2*sin(4*b*x + 4*a) - 8*b^2*sin(2*b*x + 2*a) - 4*I*b^2)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/sin(a + b*x)^3,x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*csc(a + b*x)**3, x)

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