3.852 \(\int (a+i a \tan (e+f x))^{1+m} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{-1-m} \, dx\)

Optimal. Leaf size=147 \[ \frac {a B 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{-m} \, _2F_1\left (-m,-m;1-m;\frac {1}{2} (1-i \tan (e+f x))\right )}{c f m}-\frac {(B+i A) (a+i a \tan (e+f x))^{m+1} (c-i c \tan (e+f x))^{-m-1}}{2 f (m+1)} \]

[Out]

-1/2*(I*A+B)*(a+I*a*tan(f*x+e))^(1+m)*(c-I*c*tan(f*x+e))^(-1-m)/f/(1+m)+2^m*a*B*hypergeom([-m, -m],[1-m],1/2-1
/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/c/f/m/((1+I*tan(f*x+e))^m)/((c-I*c*tan(f*x+e))^m)

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Rubi [A]  time = 0.22, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {3588, 79, 70, 69} \[ \frac {a B 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{-m} \, _2F_1\left (-m,-m;1-m;\frac {1}{2} (1-i \tan (e+f x))\right )}{c f m}-\frac {(B+i A) (a+i a \tan (e+f x))^{m+1} (c-i c \tan (e+f x))^{-m-1}}{2 f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(1 + m)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(-1 - m),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(1 + m)*(c - I*c*Tan[e + f*x])^(-1 - m))/(2*f*(1 + m)) + (2^m*a*B*Hypergeom
etric2F1[-m, -m, 1 - m, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(c*f*m*(1 + I*Tan[e + f*x])^m*(c - I
*c*Tan[e + f*x])^m)

Rule 69

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 85.98, size = 177, normalized size = 1.20 \[ \frac {a e^{i (e+2 f x)} \left (e^{i f x}\right )^m \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m (\tan (e+f x)-i) \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{-m} \sec ^{-m-1}(e+f x) (\cos (f x)+i \sin (f x))^{-m-1} (a+i a \tan (e+f x))^m \left (A+2 i B \, _2F_1\left (1,m+1;m+2;-e^{2 i (e+f x)}\right )-i B\right )}{2 c f (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(1 + m)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(-1 - m),x]

[Out]

(a*E^(I*(e + 2*f*x))*(E^(I*f*x))^m*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*(A - I*B + (2*I)*B*Hypergeome
tric2F1[1, 1 + m, 2 + m, -E^((2*I)*(e + f*x))])*Sec[e + f*x]^(-1 - m)*(Cos[f*x] + I*Sin[f*x])^(-1 - m)*(-I + T
an[e + f*x])*(a + I*a*Tan[e + f*x])^m)/(2*c*(c/(1 + E^((2*I)*(e + f*x))))^m*f*(1 + m))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A - i \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + A + i \, B\right )} \left (\frac {2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{-m - 1} e^{\left (2 i \, e m + {\left (2 i \, f m + 2 i \, f\right )} x + {\left (m + 1\right )} \log \left (\frac {a}{c}\right ) + {\left (m + 1\right )} \log \left (\frac {2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right ) + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x, algorithm="fricas")

[Out]

integral(((A - I*B)*e^(2*I*f*x + 2*I*e) + A + I*B)*(2*c/(e^(2*I*f*x + 2*I*e) + 1))^(-m - 1)*e^(2*I*e*m + (2*I*
f*m + 2*I*f)*x + (m + 1)*log(a/c) + (m + 1)*log(2*c/(e^(2*I*f*x + 2*I*e) + 1)) + 2*I*e)/(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 \[ \int {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m + 1} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{-m - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(m + 1)*(-I*c*tan(f*x + e) + c)^(-m - 1), x)

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maple [F]  time = 12.18, size = 0, normalized size = 0.00 \[ \int \left (a +i a \tan \left (f x +e \right )\right )^{1+m} \left (A +B \tan \left (f x +e \right )\right ) \left (c -i c \tan \left (f x +e \right )\right )^{-1-m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x)

[Out]

int((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(m + 1))/(c - c*tan(e + f*x)*1i)^(m + 1),x)

[Out]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(m + 1))/(c - c*tan(e + f*x)*1i)^(m + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(-1-m),x)

[Out]

Exception raised: TypeError

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