3.243 \(\int (c+d x) \csc ^3(a+b x) \sec (a+b x) \, dx\)
Optimal. Leaf size=141 \[ \frac {i d \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+\frac {(c+d x) \log (\tan (a+b x))}{b}-\frac {2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d x \log (\tan (a+b x))}{b}-\frac {d x}{2 b} \]
[Out]
-1/2*d*x/b-2*d*x*arctanh(exp(2*I*(b*x+a)))/b-1/2*d*cot(b*x+a)/b^2-1/2*(d*x+c)*cot(b*x+a)^2/b-d*x*ln(tan(b*x+a)
)/b+(d*x+c)*ln(tan(b*x+a))/b+1/2*I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2-1/2*I*d*polylog(2,exp(2*I*(b*x+a)))/b^2
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Rubi [A] time = 0.14, antiderivative size = 141, normalized size of antiderivative = 1.00,
number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used
= {2620, 14, 4420, 3473, 8, 2548, 12, 4183, 2279, 2391} \[ \frac {i d \text {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {i d \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+\frac {(c+d x) \log (\tan (a+b x))}{b}-\frac {2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d x \log (\tan (a+b x))}{b}-\frac {d x}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x],x]
[Out]
-(d*x)/(2*b) - (2*d*x*ArcTanh[E^((2*I)*(a + b*x))])/b - (d*Cot[a + b*x])/(2*b^2) - ((c + d*x)*Cot[a + b*x]^2)/
(2*b) - (d*x*Log[Tan[a + b*x]])/b + ((c + d*x)*Log[Tan[a + b*x]])/b + ((I/2)*d*PolyLog[2, -E^((2*I)*(a + b*x))
])/b^2 - ((I/2)*d*PolyLog[2, E^((2*I)*(a + b*x))])/b^2
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 14
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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 2548
Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]
Rule 2620
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Rule 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 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 4420
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Modul
e[{u = IntHide[Csc[a + b*x]^n*Sec[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)*u
, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]
Rubi steps
\begin {align*} \int (c+d x) \csc ^3(a+b x) \sec (a+b x) \, dx &=-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+\frac {(c+d x) \log (\tan (a+b x))}{b}-d \int \left (-\frac {\cot ^2(a+b x)}{2 b}+\frac {\log (\tan (a+b x))}{b}\right ) \, dx\\ &=-\frac {(c+d x) \cot ^2(a+b x)}{2 b}+\frac {(c+d x) \log (\tan (a+b x))}{b}+\frac {d \int \cot ^2(a+b x) \, dx}{2 b}-\frac {d \int \log (\tan (a+b x)) \, dx}{b}\\ &=-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x \log (\tan (a+b x))}{b}+\frac {(c+d x) \log (\tan (a+b x))}{b}-\frac {d \int 1 \, dx}{2 b}+\frac {d \int 2 b x \csc (2 a+2 b x) \, dx}{b}\\ &=-\frac {d x}{2 b}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x \log (\tan (a+b x))}{b}+\frac {(c+d x) \log (\tan (a+b x))}{b}+(2 d) \int x \csc (2 a+2 b x) \, dx\\ &=-\frac {d x}{2 b}-\frac {2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x \log (\tan (a+b x))}{b}+\frac {(c+d x) \log (\tan (a+b x))}{b}-\frac {d \int \log \left (1-e^{i (2 a+2 b x)}\right ) \, dx}{b}+\frac {d \int \log \left (1+e^{i (2 a+2 b x)}\right ) \, dx}{b}\\ &=-\frac {d x}{2 b}-\frac {2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x \log (\tan (a+b x))}{b}+\frac {(c+d x) \log (\tan (a+b x))}{b}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^2}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{2 b^2}\\ &=-\frac {d x}{2 b}-\frac {2 d x \tanh ^{-1}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {d x \log (\tan (a+b x))}{b}+\frac {(c+d x) \log (\tan (a+b x))}{b}+\frac {i d \text {Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}\\ \end {align*}
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Mathematica [A] time = 0.90, size = 210, normalized size = 1.49 \[ \frac {d \left (\frac {1}{2} i \text {Li}_2\left (-e^{2 i (a+b x)}\right )+\frac {1}{2} i (a+b x)^2-(a+b x) \log \left (1+e^{2 i (a+b x)}\right )\right )}{b^2}+\frac {d \left ((a+b x) \log \left (1-e^{2 i (a+b x)}\right )-\frac {1}{2} i \left ((a+b x)^2+\text {Li}_2\left (e^{2 i (a+b x)}\right )\right )\right )}{b^2}-\frac {d \cot (a+b x)}{2 b^2}-\frac {a d \log (\sin (a+b x))}{b^2}+\frac {a d \log (\cos (a+b x))}{b^2}-\frac {c \left (\csc ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))\right )}{2 b}-\frac {d x \csc ^2(a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)*Csc[a + b*x]^3*Sec[a + b*x],x]
[Out]
-1/2*(d*Cot[a + b*x])/b^2 - (d*x*Csc[a + b*x]^2)/(2*b) + (a*d*Log[Cos[a + b*x]])/b^2 - (c*(Csc[a + b*x]^2 + 2*
Log[Cos[a + b*x]] - 2*Log[Sin[a + b*x]]))/(2*b) - (a*d*Log[Sin[a + b*x]])/b^2 + (d*((I/2)*(a + b*x)^2 - (a + b
*x)*Log[1 + E^((2*I)*(a + b*x))] + (I/2)*PolyLog[2, -E^((2*I)*(a + b*x))]))/b^2 + (d*((a + b*x)*Log[1 - E^((2*
I)*(a + b*x))] - (I/2)*((a + b*x)^2 + PolyLog[2, E^((2*I)*(a + b*x))])))/b^2
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fricas [B] time = 0.54, size = 942, normalized size = 6.68 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x, algorithm="fricas")
[Out]
1/2*(b*d*x + d*cos(b*x + a)*sin(b*x + a) + b*c + (-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)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(I*co
s(b*x + a) + sin(b*x + a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) + (I*d*cos(b*x +
a)^2 - I*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(-I*cos(b*x + a) - 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*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(cos(b*x + a) + I*sin(b*x + a) + I) - (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*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*
log(cos(b*x + a) - I*sin(b*x + a) + I) + (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(I*cos(b*x + a) + sin
(b*x + a) + 1) + (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b*d*x
- (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d*x - (b*d*x + a*d)*cos(b*x
+ a)^2 + a*d)*log(-I*cos(b*x + a) - 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*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (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*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-cos(
b*x + a) - I*sin(b*x + a) + I))/(b^2*cos(b*x + a)^2 - b^2)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \csc \left (b x + a\right )^{3} \sec \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x, algorithm="giac")
[Out]
integrate((d*x + c)*csc(b*x + a)^3*sec(b*x + a), x)
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maple [B] time = 0.12, size = 270, normalized size = 1.91 \[ \frac {2 b d x \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b c \,{\mathrm e}^{2 i \left (b x +a \right )}-i d \,{\mathrm e}^{2 i \left (b x +a \right )}+i d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {c \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {d \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b}+\frac {i d \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {i d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {i d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)*csc(b*x+a)^3*sec(b*x+a),x)
[Out]
(2*b*d*x*exp(2*I*(b*x+a))+2*b*c*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))+I*d)/b^2/(exp(2*I*(b*x+a))-1)^2+1/b*c*ln
(exp(I*(b*x+a))-1)-1/b*c*ln(1+exp(2*I*(b*x+a)))+1/b*c*ln(exp(I*(b*x+a))+1)-1/b*d*ln(1+exp(2*I*(b*x+a)))*x+1/2*
I*d*polylog(2,-exp(2*I*(b*x+a)))/b^2+1/b*d*ln(exp(I*(b*x+a))+1)*x-I*d*polylog(2,-exp(I*(b*x+a)))/b^2+1/b*d*ln(
1-exp(I*(b*x+a)))*x+1/b^2*d*ln(1-exp(I*(b*x+a)))*a-I*d*polylog(2,exp(I*(b*x+a)))/b^2-1/b^2*d*a*ln(exp(I*(b*x+a
))-1)
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maxima [B] time = 0.62, size = 1035, normalized size = 7.34 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)^3*sec(b*x+a),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(2*b*x + 2*a), cos(2*b*x + 2*a) + 1)
- (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)*ar
ctan2(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 + 2*d)*cos(2*b*x + 2*a) - (d*cos(4*b*x + 4*a) - 2*d*cos(2*b*x + 2*a) + I*d*sin(4*b*x + 4*a) - 2*I*d*sin(
2*b*x + 2*a) + d)*dilog(-e^(2*I*b*x + 2*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)) + (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(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*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) + (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)*si
n(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 + 4*b*c - 2*I*d)*sin(2*b*x + 2*a) - 2*d)/(-2*I*b^2*cos(4*b*x + 4*a) + 4*I*b^2*cos(2*b*x + 2*a) + 2*b
^2*sin(4*b*x + 4*a) - 4*b^2*sin(2*b*x + 2*a) - 2*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)/(cos(a + b*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 \[ \int \left (c + d x\right ) \csc ^{3}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)*csc(b*x+a)**3*sec(b*x+a),x)
[Out]
Integral((c + d*x)*csc(a + b*x)**3*sec(a + b*x), x)
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