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

Optimal. Leaf size=43 \[ \frac {a (e \tan (c+d x))^{m+1} \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d e (m+1)} \]

[Out]

a*hypergeom([1, 1+m],[2+m],I*tan(d*x+c))*(e*tan(d*x+c))^(1+m)/d/e/(1+m)

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3537, 64} \[ \frac {a (e \tan (c+d x))^{m+1} \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 3537

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

Rubi steps

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

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Mathematica [B]  time = 1.03, size = 159, normalized size = 3.70 \[ \frac {a e^{-i c} 2^{-m-1} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{m+1} \cos (c+d x) (1+i \tan (c+d x)) \, _2F_1\left (m+1,m+1;m+2;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right ) \tan ^{-m}(c+d x) (e \tan (c+d x))^m}{d (m+1) (\cos (d x)+i \sin (d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a \left (\frac {-i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \left (e \tan \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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