3.318 \(\int (d \tan (e+f x))^n (a-i a \tan (e+f x)) \, dx\)

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

[Out]

a*hypergeom([1, 1+n],[2+n],-I*tan(f*x+e))*(d*tan(f*x+e))^(1+n)/d/f/(1+n)

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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} \[ \frac {a (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;-i \tan (e+f x))}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.07, size = 44, normalized size = 1.02 \[ \frac {a \tan (e+f x) (d \tan (e+f x))^n \, _2F_1(1,n+1;n+2;-i \tan (e+f x))}{f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a \left (\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-I*a*tan(f*x + e) + a)*(d*tan(f*x + e))^n, x)

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maple [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (f x +e \right )\right )^{n} \left (a -i a \tan \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a-I*a*tan(f*x+e)),x)

[Out]

int((d*tan(f*x+e))^n*(a-I*a*tan(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-I*a*tan(f*x + e) + a)*(d*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^n*(a - a*tan(e + f*x)*1i),x)

[Out]

int((d*tan(e + f*x))^n*(a - a*tan(e + f*x)*1i), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n*(a-I*a*tan(f*x+e)),x)

[Out]

-I*a*(Integral(I*(d*tan(e + f*x))**n, x) + Integral((d*tan(e + f*x))**n*tan(e + f*x), x))

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