3.1181 \(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=119 \[ -\frac{i (c-i d)^2 (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m}+\frac{2 c d (a+i a \tan (e+f x))^m}{f m}-\frac{i d^2 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.166736, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3543, 3527, 3481, 68} \[ -\frac{i (c-i d)^2 (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m}+\frac{2 c d (a+i a \tan (e+f x))^m}{f m}-\frac{i d^2 (a+i a \tan (e+f x))^{m+1}}{a f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

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

Mathematica [B]  time = 17.136, size = 246, normalized size = 2.07 \[ -\frac{i 2^{m-1} \left (e^{i f x}\right )^m e^{-i (e m+e+2 f m x+f x)} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{m+1} \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m \left (\frac{2 \left (c^2+d^2\right ) e^{i (e (m+2)+2 f (m+1) x)}}{m+1}+\frac{(c-i d)^2 e^{i (e (m+4)+2 f (m+2) x)} \, _2F_1\left (1,1;m+3;-e^{2 i (e+f x)}\right )}{m+2}+\frac{(c+i d)^2 e^{i m (e+2 f x)} \left (e^{2 i (e+f x)}+m+1\right )}{m (m+1)}\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(-1 + m)*(E^(I*f*x))^m*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(1 + m)*((2*(c^2 + d^2)*E^(I*(e*(2
+ m) + 2*f*(1 + m)*x)))/(1 + m) + ((c + I*d)^2*E^(I*m*(e + 2*f*x))*(1 + E^((2*I)*(e + f*x)) + m))/(m*(1 + m))
+ ((c - I*d)^2*E^(I*(e*(4 + m) + 2*f*(2 + m)*x))*Hypergeometric2F1[1, 1, 3 + m, -E^((2*I)*(e + f*x))])/(2 + m)
)*(a + I*a*Tan[e + f*x])^m)/(E^(I*(e + e*m + f*x + 2*f*m*x))*f*Sec[e + f*x]^m*(Cos[f*x] + I*Sin[f*x])^m)

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Maple [F]  time = 0.501, 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 ) ^{2}\, 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))^2,x)

[Out]

int((a+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^2,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 )}^{2}{\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))^2,x, algorithm="maxima")

[Out]

integrate((d*tan(f*x + e) + c)^2*(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 (c^{2} + 2 i \, c d - d^{2} +{\left (c^{2} - 2 i \, c d - d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \,{\left (c^{2} + d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\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 (4 i \, f x + 4 i \, e\right )} + 2 \, 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))^2,x, algorithm="fricas")

[Out]

integral((c^2 + 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*e^(4*I*f*x + 4*I*e) + 2*(c^2 + d^2)*e^(2*I*f*x + 2*I*e))
*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m/(e^(4*I*f*x + 4*I*e) + 2*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 )^{2}\, 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))**2,x)

[Out]

Integral((a*(I*tan(e + f*x) + 1))**m*(c + d*tan(e + f*x))**2, 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 )}^{2}{\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))^2,x, algorithm="giac")

[Out]

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