3.1.37 \(\int \frac {(c+d x)^m}{(a+i a \tan (e+f x))^2} \, dx\) [37]

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

[Out]

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

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Rubi [A]
time = 0.14, 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 = {3810, 2212} \begin {gather*} \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)} \end {gather*}

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 2212

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

Rule 3810

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/b)*(e + f*x))/(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))^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 = 151.50, size = 205, normalized size = 1.20 \begin {gather*} \frac {4^{-2-m} e^{-2 i e} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \left (4^{1+m} e^{4 i e} f (c+d x) \left (\frac {i f (c+d x)}{d}\right )^m+i 2^{2+m} d e^{\frac {2 i (d e+c f)}{d}} (1+m) \text {Gamma}\left (1+m,\frac {2 i f (c+d x)}{d}\right )+i d e^{\frac {4 i c f}{d}} (1+m) \text {Gamma}\left (1+m,\frac {4 i f (c+d x)}{d}\right )\right ) \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2}{d f (1+m) (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{m}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 - 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 = 0.10, size = 146, normalized size = 0.85 \begin {gather*} -\frac {4 \, {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) - 2 i \, c f + 2 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - {\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, f}{d}\right ) - 4 i \, c f + 4 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, d f x - i \, c f\right )}}{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 {gather*}

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*(4*(-I*d*m - I*d)*e^(-(d*m*log(2*I*f/d) - 2*I*c*f + 2*I*d*e)/d)*gamma(m + 1, -2*(-I*d*f*x - I*c*f)/d) -
(I*d*m + I*d)*e^(-(d*m*log(4*I*f/d) - 4*I*c*f + 4*I*d*e)/d)*gamma(m + 1, -4*(-I*d*f*x - 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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (c + d x\right )^{m}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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