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

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

[Out]

1/8*(d*x+c)^(1+m)/a^3/d/(1+m)+3*I*2^(-4-m)*(d*x+c)^m*GAMMA(1+m,2*I*f*(d*x+c)/d)/a^3/exp(2*I*(e-c*f/d))/f/((I*f
*(d*x+c)/d)^m)+3*I*2^(-5-2*m)*(d*x+c)^m*GAMMA(1+m,4*I*f*(d*x+c)/d)/a^3/exp(4*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m
)+I*2^(-4-m)*3^(-1-m)*(d*x+c)^m*GAMMA(1+m,6*I*f*(d*x+c)/d)/a^3/exp(6*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)

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Rubi [A]  time = 0.24, antiderivative size = 251, normalized size of antiderivative = 1.00, 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 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 \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*}

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Mathematica [A]  time = 68.70, size = 269, normalized size = 1.07 \[ \frac {e^{-3 i e} 2^{-2 m-5} 3^{-m-1} (c+d x)^m \sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac {i f (c+d x)}{d}\right )^{-m} \left (e^{6 i e} f 12^{m+1} (c+d x) \left (\frac {i f (c+d x)}{d}\right )^m+i d 2^{m+1} 3^{m+2} (m+1) e^{2 i \left (\frac {c f}{d}+2 e\right )} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )+i d 3^{m+2} (m+1) e^{\frac {4 i c f}{d}+2 i e} \Gamma \left (m+1,\frac {4 i f (c+d x)}{d}\right )+i d 2^{m+1} (m+1) e^{\frac {6 i c f}{d}} \Gamma \left (m+1,\frac {6 i f (c+d x)}{d}\right )\right )}{d f (m+1) (a+i a \tan (e+f x))^3} \]

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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fricas [A]  time = 0.62, size = 192, normalized size = 0.76 \[ \frac {{\left (2 i \, d m + 2 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {6 i \, f}{d}\right ) + 6 i \, d e - 6 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {6 i \, d f x + 6 i \, c f}{d}\right ) + {\left (9 i \, d m + 9 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, f}{d}\right ) + 4 i \, d e - 4 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {4 i \, d f x + 4 i \, c f}{d}\right ) + {\left (18 i \, d m + 18 i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, d f x + 2 i \, c f}{d}\right ) + 12 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{96 \, {\left (a^{3} d f m + a^{3} d f\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{m}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

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)

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{m}}{\left (a +i a \tan \left (f x +e \right )\right )^{3}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (6 \, f x + 6 \, e\right )\,{d x} + 3 \, {\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (4 \, f x + 4 \, e\right )\,{d x} + 3 \, {\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} - {\left (i \, d m + i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (6 \, f x + 6 \, e\right )\,{d x} - {\left (3 i \, d m + 3 i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (4 \, f x + 4 \, e\right )\,{d x} - {\left (3 i \, d m + 3 i \, d\right )} \int {\left (d x + c\right )}^{m} \sin \left (2 \, f x + 2 \, e\right )\,{d x} + e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{8 \, {\left (a^{3} d m + a^{3} d\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^m/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

int((c + d*x)^m/(a + a*tan(e + f*x)*1i)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\left (c + d x\right )^{m}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]

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]

I*Integral((c + d*x)**m/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x)/a**3

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