3.454 \(\int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=86 \[ \frac {i 2^{\frac {m}{2}-1} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \, _2F_1\left (2-\frac {m}{2},\frac {m}{2};\frac {m+2}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{a d m} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3505, 3523, 70, 69} \[ \frac {i 2^{\frac {m}{2}-1} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \text {Hypergeometric2F1}\left (2-\frac {m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{a d m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3505

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

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.70, size = 150, normalized size = 1.74 \[ -\frac {i 2^{m-1} e^{-i (c+2 d x)} \left (1+e^{2 i (c+d x)}\right )^2 \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m (\cos (d x)+i \sin (d x)) \, _2F_1\left (1,1-\frac {m}{2};\frac {m}{2};-e^{2 i (c+d x)}\right ) \sec ^{1-m}(c+d x) (e \sec (c+d x))^m}{d (m-2) (a+i a \tan (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.32, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{a +i a \tan \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),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((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c)),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 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e/cos(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 \[ - \frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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