3.1177 \(\int \frac {(c+d \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=193 \[ \frac {(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {c+d \tan (e+f x)}{c-i d}\right )}{4 a f (n+1) (d+i c)}+\frac {(i c+2 d n-d) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {c+d \tan (e+f x)}{c+i d}\right )}{4 a f (n+1) (c+i d)^2}-\frac {(c+d \tan (e+f x))^{n+1}}{2 f (-d+i c) (a+i a \tan (e+f x))} \]

[Out]

1/4*hypergeom([1, 1+n],[2+n],(c+d*tan(f*x+e))/(c-I*d))*(c+d*tan(f*x+e))^(1+n)/a/(I*c+d)/f/(1+n)+1/4*(I*c-d+2*d
*n)*hypergeom([1, 1+n],[2+n],(c+d*tan(f*x+e))/(c+I*d))*(c+d*tan(f*x+e))^(1+n)/a/(c+I*d)^2/f/(1+n)-1/2*(c+d*tan
(f*x+e))^(1+n)/(I*c-d)/f/(a+I*a*tan(f*x+e))

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Rubi [A]  time = 0.28, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3552, 3539, 3537, 68} \[ \frac {(c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {c+d \tan (e+f x)}{c-i d}\right )}{4 a f (n+1) (d+i c)}+\frac {(i c+2 d n-d) (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {c+d \tan (e+f x)}{c+i d}\right )}{4 a f (n+1) (c+i d)^2}-\frac {(c+d \tan (e+f x))^{n+1}}{2 f (-d+i c) (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, 1 + n, 2 + n, (c + d*Tan[e + f*x])/(c - I*d)]*(c + d*Tan[e + f*x])^(1 + n))/(4*a*(I*c +
d)*f*(1 + n)) + ((I*c - d + 2*d*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (c + d*Tan[e + f*x])/(c + I*d)]*(c + d*T
an[e + f*x])^(1 + n))/(4*a*(c + I*d)^2*f*(1 + n)) - (c + d*Tan[e + f*x])^(1 + n)/(2*(I*c - d)*f*(a + I*a*Tan[e
+ f*x]))

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]

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]

Rule 3539

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

Rule 3552

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

Rubi steps

\begin {align*} \int \frac {(c+d \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx &=-\frac {(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac {\int (c+d \tan (e+f x))^n (a (i c-d (1-n))-i a d n \tan (e+f x)) \, dx}{2 a^2 (i c-d)}\\ &=-\frac {(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac {\int (1+i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{4 a}+\frac {(c+i d (1-2 n)) \int (1-i \tan (e+f x)) (c+d \tan (e+f x))^n \, dx}{4 a (c+i d)}\\ &=-\frac {(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}+\frac {i \operatorname {Subst}\left (\int \frac {(c-i d x)^n}{-1+x} \, dx,x,i \tan (e+f x)\right )}{4 a f}-\frac {(i (c+i d (1-2 n))) \operatorname {Subst}\left (\int \frac {(c+i d x)^n}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{4 a (c+i d) f}\\ &=\frac {\, _2F_1\left (1,1+n;2+n;\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{4 a (i c+d) f (1+n)}+\frac {(i c-d (1-2 n)) \, _2F_1\left (1,1+n;2+n;\frac {c+d \tan (e+f x)}{c+i d}\right ) (c+d \tan (e+f x))^{1+n}}{4 a (c+i d)^2 f (1+n)}-\frac {(c+d \tan (e+f x))^{1+n}}{2 (i c-d) f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [F]  time = 30.70, size = 0, normalized size = 0.00 \[ \int \frac {(c+d \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.49, size = 0, normalized size = 0.00 \[ \int \frac {\left (c +d \tan \left (f x +e \right )\right )^{n}}{a +i a \tan \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c + 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 \[ - \frac {i \int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{n}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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