3.222 \(\int (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=78 \[ \frac{B (a+i a \tan (c+d x))^n}{d n}-\frac{(B+i A) (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (1+i \tan (c+d x))\right )}{2 d n} \]

[Out]

(B*(a + I*a*Tan[c + d*x])^n)/(d*n) - ((I*A + B)*Hypergeometric2F1[1, n, 1 + n, (1 + I*Tan[c + d*x])/2]*(a + I*
a*Tan[c + d*x])^n)/(2*d*n)

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Rubi [A]  time = 0.0656323, 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} \[ \frac{B (a+i a \tan (c+d x))^n}{d n}-\frac{(B+i A) (a+i a \tan (c+d x))^n \, _2F_1\left (1,n;n+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

(B*(a + I*a*Tan[c + d*x])^n)/(d*n) - ((I*A + B)*Hypergeometric2F1[1, n, 1 + n, (1 + I*Tan[c + d*x])/2]*(a + I*
a*Tan[c + d*x])^n)/(2*d*n)

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

Mathematica [A]  time = 7.15504, size = 152, normalized size = 1.95 \[ \frac{2^{n-1} \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n \left ((n+1) (B-i A)-i n (A-i B) e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (1,1,n+2,-e^{2 i (c+d x)}\right )\right )}{d n (n+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

(2^(-1 + n)*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(((-I)*A + B)*(1 + n) - I*(A - I*B)*E^
((2*I)*(c + d*x))*n*Hypergeometric2F1[1, 1, 2 + n, -E^((2*I)*(c + d*x))])*(a + I*a*Tan[c + d*x])^n)/(d*n*(1 +
n)*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]  time = 0.787, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(((A - I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n/(e^(2
*I*d*x + 2*I*c) + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((a*(I*tan(c + d*x) + 1))**n*(A + B*tan(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n, x)