3.1182 \(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x)) \, dx\)

Optimal. Leaf size=78 \[ \frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(d+i c) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]

[Out]

(d*(a + I*a*Tan[e + f*x])^m)/(f*m) - ((I*c + d)*Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*
a*Tan[e + f*x])^m)/(2*f*m)

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Rubi [A]  time = 0.0697399, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3527, 3481, 68} \[ \frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(d+i c) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a + I*a*Tan[e + f*x])^m)/(f*m) - ((I*c + d)*Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*
a*Tan[e + f*x])^m)/(2*f*m)

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x)) \, dx &=\frac{d (a+i a \tan (e+f x))^m}{f m}-(-c+i d) \int (a+i a \tan (e+f x))^m \, dx\\ &=\frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(a (i c+d)) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{d (a+i a \tan (e+f x))^m}{f m}-\frac{(i c+d) \, _2F_1\left (1,m;1+m;\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 f m}\\ \end{align*}

Mathematica [A]  time = 5.78361, size = 152, normalized size = 1.95 \[ \frac{2^{m-1} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m \left ((m+1) (d-i c)-i m (c-i d) e^{2 i (e+f x)} \, _2F_1\left (1,1;m+2;-e^{2 i (e+f x)}\right )\right )}{f m (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + m)*(E^(I*f*x))^m*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*(((-I)*c + d)*(1 + m) - I*(c - I*d)*E^
((2*I)*(e + f*x))*m*Hypergeometric2F1[1, 1, 2 + m, -E^((2*I)*(e + f*x))])*(a + I*a*Tan[e + f*x])^m)/(f*m*(1 +
m)*Sec[e + f*x]^m*(Cos[f*x] + I*Sin[f*x])^m)

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Maple [F]  time = 0.771, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d\right )} \left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{m} \left (c + d \tan{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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