3.35 \(\int \frac {(c+d x)^m}{(a+i a \cot (e+f x))^2} \, dx\)

Optimal. Leaf size=171 \[ \frac {i 2^{-m-2} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{a^2 f}-\frac {i 4^{-m-2} e^{4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i f (c+d x)}{d}\right )}{a^2 f}+\frac {(c+d x)^{m+1}}{4 a^2 d (m+1)} \]

[Out]

1/4*(d*x+c)^(1+m)/a^2/d/(1+m)+I*2^(-2-m)*exp(2*I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-2*I*f*(d*x+c)/d)/a^2/f/((-I*f
*(d*x+c)/d)^m)-I*4^(-2-m)*exp(4*I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-4*I*f*(d*x+c)/d)/a^2/f/((-I*f*(d*x+c)/d)^m)

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Rubi [A]  time = 0.18, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3729, 2181} \[ \frac {i 2^{-m-2} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{a^2 f}-\frac {i 4^{-m-2} e^{4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {4 i f (c+d x)}{d}\right )}{a^2 f}+\frac {(c+d x)^{m+1}}{4 a^2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m/(a + I*a*Cot[e + f*x])^2,x]

[Out]

(c + d*x)^(1 + m)/(4*a^2*d*(1 + m)) + (I*2^(-2 - m)*E^((2*I)*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*I)*f
*(c + d*x))/d])/(a^2*f*(((-I)*f*(c + d*x))/d)^m) - (I*4^(-2 - m)*E^((4*I)*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 +
m, ((-4*I)*f*(c + d*x))/d])/(a^2*f*(((-I)*f*(c + d*x))/d)^m)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3729

Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c
+ d*x)^m, (1/(2*a) + E^((2*a*(e + f*x))/b)/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
+ b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(c+d x)^m}{(a+i a \cot (e+f x))^2} \, dx &=\int \left (\frac {(c+d x)^m}{4 a^2}-\frac {e^{2 i e+2 i f x} (c+d x)^m}{2 a^2}+\frac {e^{4 i e+4 i f x} (c+d x)^m}{4 a^2}\right ) \, dx\\ &=\frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac {\int e^{4 i e+4 i f x} (c+d x)^m \, dx}{4 a^2}-\frac {\int e^{2 i e+2 i f x} (c+d x)^m \, dx}{2 a^2}\\ &=\frac {(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac {i 2^{-2-m} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{a^2 f}-\frac {i 4^{-2-m} e^{4 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i f (c+d x)}{d}\right )}{a^2 f}\\ \end {align*}

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Mathematica [A]  time = 2.83, size = 152, normalized size = 0.89 \[ \frac {(c+d x)^m \left (i 2^{2-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )-i 4^{-m} e^{4 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i f (c+d x)}{d}\right )+\frac {4 f (c+d x)}{d (m+1)}\right )}{16 a^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m/(a + I*a*Cot[e + f*x])^2,x]

[Out]

((c + d*x)^m*((4*f*(c + d*x))/(d*(1 + m)) + (I*2^(2 - m)*E^((2*I)*(e - (c*f)/d))*Gamma[1 + m, ((-2*I)*f*(c + d
*x))/d])/(((-I)*f*(c + d*x))/d)^m - (I*E^((4*I)*(e - (c*f)/d))*Gamma[1 + m, ((-4*I)*f*(c + d*x))/d])/(4^m*(((-
I)*f*(c + d*x))/d)^m)))/(16*a^2*f)

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fricas [A]  time = 0.68, size = 140, normalized size = 0.82 \[ \frac {{\left (4 i \, d m + 4 i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, d f x - 2 i \, c f}{d}\right ) + {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {4 i \, f}{d}\right ) - 4 i \, d e + 4 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-4 i \, d f x - 4 i \, c f}{d}\right ) + 4 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{16 \, {\left (a^{2} d f m + a^{2} d f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m/(a+I*a*cot(f*x+e))^2,x, algorithm="fricas")

[Out]

1/16*((4*I*d*m + 4*I*d)*e^(-(d*m*log(-2*I*f/d) - 2*I*d*e + 2*I*c*f)/d)*gamma(m + 1, (-2*I*d*f*x - 2*I*c*f)/d)
+ (-I*d*m - I*d)*e^(-(d*m*log(-4*I*f/d) - 4*I*d*e + 4*I*c*f)/d)*gamma(m + 1, (-4*I*d*f*x - 4*I*c*f)/d) + 4*(d*
f*x + c*f)*(d*x + c)^m)/(a^2*d*f*m + a^2*d*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \cot \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m/(a+I*a*cot(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^m/(I*a*cot(f*x + e) + a)^2, x)

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maple [F]  time = 2.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{m}}{\left (a +i a \cot \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m/(a+I*a*cot(f*x+e))^2,x)

[Out]

int((d*x+c)^m/(a+I*a*cot(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (4 \, f x + 4 \, e\right )\,{d x} - 2 \, {\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} + {\left (i \, d m + i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (4 \, f x + 4 \, e\right )\,{d x} + {\left (-2 i \, d m - 2 i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (2 \, f x + 2 \, e\right )\,{d x} + e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{4 \, {\left (a^{2} d m + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m/(a+I*a*cot(f*x+e))^2,x, algorithm="maxima")

[Out]

1/4*((d*m + d)*integrate((d*x + c)^m*cos(4*f*x + 4*e), x) - 2*(d*m + d)*integrate((d*x + c)^m*cos(2*f*x + 2*e)
, x) + (I*d*m + I*d)*integrate((d*x + c)^m*sin(4*f*x + 4*e), x) + (-2*I*d*m - 2*I*d)*integrate((d*x + c)^m*sin
(2*f*x + 2*e), x) + e^(m*log(d*x + c) + log(d*x + c)))/(a^2*d*m + a^2*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {cot}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^m/(a + a*cot(e + f*x)*1i)^2,x)

[Out]

int((c + d*x)^m/(a + a*cot(e + f*x)*1i)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\left (c + d x\right )^{m}}{\cot ^{2}{\left (e + f x \right )} - 2 i \cot {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m/(a+I*a*cot(f*x+e))**2,x)

[Out]

-Integral((c + d*x)**m/(cot(e + f*x)**2 - 2*I*cot(e + f*x) - 1), x)/a**2

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