### 3.37 $$\int \frac{(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.184187, antiderivative size = 171, normalized size of antiderivative = 1., 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*Tan[e + f*x])^2,x]

[Out]

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

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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^m}{(a+i a \tan (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*}

Mathematica [A]  time = 38.0859, size = 192, normalized size = 1.12 $\frac{(c+d x)^m \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (i 4^{-m} e^{\frac{4 i c f}{d}-2 i e} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i f (c+d x)}{d}\right )+i 2^{2-m} e^{\frac{2 i c f}{d}} \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )+\frac{4 e^{2 i e} f (c+d x)}{d (m+1)}\right )}{16 f (a+i a \tan (e+f x))^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{m}}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m/(a+I*a*tan(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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Fricas [A]  time = 1.67203, size = 354, normalized size = 2.07 \begin{align*} \frac{{\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 ) +{\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 ) + 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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((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*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) + 4*(d*f*x +
c*f)*(d*x + c)^m)/(a^2*d*f*m + a^2*d*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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