3.5.54 \(\int \frac {(e \sec (c+d x))^m}{a+i a \tan (c+d x)} \, dx\) [454]

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

[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 = 0.13, 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 = {3586, 3604, 72, 71} \begin {gather*} \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} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 3586

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 3604

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 ) \text {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 ) \text {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 [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(212\) vs. \(2(86)=172\).
time = 1.02, size = 212, normalized size = 2.47 \begin {gather*} -\frac {i 2^{-1+m} e^{-i (c+d m x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (1+e^{2 i (c+d x)}\right )^m \left (e^{i d (-2+m) x} m \, _2F_1\left (\frac {1}{2} (-2+m),m;\frac {m}{2};-e^{2 i (c+d x)}\right )+e^{i (2 c+d m x)} (-2+m) \, _2F_1\left (\frac {m}{2},m;\frac {2+m}{2};-e^{2 i (c+d x)}\right )\right ) \sec ^{1-m}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))}{d (-2+m) m (a+i a \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[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)))^m*(E^(I*d*(-2 + m)*x)
*m*Hypergeometric2F1[(-2 + m)/2, m, m/2, -E^((2*I)*(c + d*x))] + E^(I*(2*c + d*m*x))*(-2 + m)*Hypergeometric2F
1[m/2, 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 + d*m*x))*(-2 + m)*m*(a + I*a*Tan[c + d*x]))

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Maple [F]
time = 0.94, 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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^(I*d*x + I*c + 1)/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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