3.38 $$\int \frac{(c+d x)^m}{(a+i a \tan (e+f x))^3} \, dx$$

Optimal. Leaf size=251 $\frac{3 i 2^{-m-4} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{a^3 f}+\frac{3 i 2^{-2 m-5} e^{-4 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i f (c+d x)}{d}\right )}{a^3 f}+\frac{i 2^{-m-4} 3^{-m-1} e^{-6 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{6 i f (c+d x)}{d}\right )}{a^3 f}+\frac{(c+d x)^{m+1}}{8 a^3 d (m+1)}$

[Out]

(c + d*x)^(1 + m)/(8*a^3*d*(1 + m)) + ((3*I)*2^(-4 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(a^3*
E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + ((3*I)*2^(-5 - 2*m)*(c + d*x)^m*Gamma[1 + m, ((4*I)*f*(c +
d*x))/d])/(a^3*E^((4*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (I*2^(-4 - m)*3^(-1 - m)*(c + d*x)^m*Gamma[1
+ m, ((6*I)*f*(c + d*x))/d])/(a^3*E^((6*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)

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Rubi [A]  time = 0.241977, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {3729, 2181} $\frac{3 i 2^{-m-4} e^{-2 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )}{a^3 f}+\frac{3 i 2^{-2 m-5} e^{-4 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i f (c+d x)}{d}\right )}{a^3 f}+\frac{i 2^{-m-4} 3^{-m-1} e^{-6 i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{6 i f (c+d x)}{d}\right )}{a^3 f}+\frac{(c+d x)^{m+1}}{8 a^3 d (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c + d*x)^m/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(c + d*x)^(1 + m)/(8*a^3*d*(1 + m)) + ((3*I)*2^(-4 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(a^3*
E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + ((3*I)*2^(-5 - 2*m)*(c + d*x)^m*Gamma[1 + m, ((4*I)*f*(c +
d*x))/d])/(a^3*E^((4*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) + (I*2^(-4 - m)*3^(-1 - m)*(c + d*x)^m*Gamma[1
+ m, ((6*I)*f*(c + d*x))/d])/(a^3*E^((6*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))^3} \, dx &=\int \left (\frac{(c+d x)^m}{8 a^3}+\frac{3 e^{-2 i e-2 i f x} (c+d x)^m}{8 a^3}+\frac{3 e^{-4 i e-4 i f x} (c+d x)^m}{8 a^3}+\frac{e^{-6 i e-6 i f x} (c+d x)^m}{8 a^3}\right ) \, dx\\ &=\frac{(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac{\int e^{-6 i e-6 i f x} (c+d x)^m \, dx}{8 a^3}+\frac{3 \int e^{-2 i e-2 i f x} (c+d x)^m \, dx}{8 a^3}+\frac{3 \int e^{-4 i e-4 i f x} (c+d x)^m \, dx}{8 a^3}\\ &=\frac{(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac{3 i 2^{-4-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^3 f}+\frac{3 i 2^{-5-2 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^3 f}+\frac{i 2^{-4-m} 3^{-1-m} e^{-6 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{6 i f (c+d x)}{d}\right )}{a^3 f}\\ \end{align*}

Mathematica [A]  time = 55.0155, size = 269, normalized size = 1.07 $\frac{e^{-3 i e} 2^{-2 m-5} 3^{-m-1} (c+d x)^m \sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{i f (c+d x)}{d}\right )^{-m} \left (i d 2^{m+1} 3^{m+2} (m+1) e^{2 i \left (\frac{c f}{d}+2 e\right )} \text{Gamma}\left (m+1,\frac{2 i f (c+d x)}{d}\right )+i d 3^{m+2} (m+1) e^{\frac{4 i c f}{d}+2 i e} \text{Gamma}\left (m+1,\frac{4 i f (c+d x)}{d}\right )+i d 2^{m+1} (m+1) e^{\frac{6 i c f}{d}} \text{Gamma}\left (m+1,\frac{6 i f (c+d x)}{d}\right )+e^{6 i e} f 12^{m+1} (c+d x) \left (\frac{i f (c+d x)}{d}\right )^m\right )}{d f (m+1) (a+i a \tan (e+f x))^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c + d*x)^m/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(2^(-5 - 2*m)*3^(-1 - m)*(c + d*x)^m*(12^(1 + m)*E^((6*I)*e)*f*(c + d*x)*((I*f*(c + d*x))/d)^m + I*2^(1 + m)*3
^(2 + m)*d*E^((2*I)*(2*e + (c*f)/d))*(1 + m)*Gamma[1 + m, ((2*I)*f*(c + d*x))/d] + I*3^(2 + m)*d*E^((2*I)*e +
((4*I)*c*f)/d)*(1 + m)*Gamma[1 + m, ((4*I)*f*(c + d*x))/d] + I*2^(1 + m)*d*E^(((6*I)*c*f)/d)*(1 + m)*Gamma[1 +
m, ((6*I)*f*(c + d*x))/d])*Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3)/(d*E^((3*I)*e)*f*(1 + m)*((I*f*(c + d*x)
)/d)^m*(a + I*a*Tan[e + f*x])^3)

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Maple [F]  time = 0.136, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{m}}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{3}}}\, 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))^3,x)

[Out]

int((d*x+c)^m/(a+I*a*tan(f*x+e))^3,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 (6 \, f x + 6 \, e\right )\,{d x} + 3 \,{\left (d m + d\right )} \int{\left (d x + c\right )}^{m} \cos \left (4 \, f x + 4 \, e\right )\,{d x} + 3 \,{\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 (6 \, f x + 6 \, e\right )\,{d x} -{\left (3 i \, d m + 3 i \, d\right )} \int{\left (d x + c\right )}^{m} \sin \left (4 \, f x + 4 \, e\right )\,{d x} -{\left (3 i \, d m + 3 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 )}}{8 \,{\left (a^{3} d m + a^{3} 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))^3,x, algorithm="maxima")

[Out]

1/8*((d*m + d)*integrate((d*x + c)^m*cos(6*f*x + 6*e), x) + 3*(d*m + d)*integrate((d*x + c)^m*cos(4*f*x + 4*e)
, x) + 3*(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(6*f*x
+ 6*e), x) - (3*I*d*m + 3*I*d)*integrate((d*x + c)^m*sin(4*f*x + 4*e), x) - (3*I*d*m + 3*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^3*d*m + a^3*d)

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Fricas [A]  time = 1.74322, size = 504, normalized size = 2.01 \begin{align*} \frac{{\left (2 i \, d m + 2 i \, d\right )} e^{\left (-\frac{d m \log \left (\frac{6 i \, f}{d}\right ) + 6 i \, d e - 6 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{6 i \, d f x + 6 i \, c f}{d}\right ) +{\left (9 i \, d m + 9 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 (18 i \, d m + 18 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 ) + 12 \,{\left (d f x + c f\right )}{\left (d x + c\right )}^{m}}{96 \,{\left (a^{3} d f m + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/96*((2*I*d*m + 2*I*d)*e^(-(d*m*log(6*I*f/d) + 6*I*d*e - 6*I*c*f)/d)*gamma(m + 1, (6*I*d*f*x + 6*I*c*f)/d) +
(9*I*d*m + 9*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) + (18*I*
d*m + 18*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) + 12*(d*f*x
+ c*f)*(d*x + c)^m)/(a^3*d*f*m + a^3*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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^m/(I*a*tan(f*x + e) + a)^3, x)