3.462 \(\int \frac{(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac{i 2^{\frac{m-3}{2}} (1+i \tan (c+d x))^{\frac{1-m}{2}} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{5-m}{2},\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{a d m \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(I*2^((-3 + m)/2)*Hypergeometric2F1[(5 - m)/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m*(1 +
I*Tan[c + d*x])^((1 - m)/2))/(a*d*m*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.208719, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ \frac{i 2^{\frac{m-3}{2}} (1+i \tan (c+d x))^{\frac{1-m}{2}} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{5-m}{2},\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{a d m \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(I*2^((-3 + m)/2)*Hypergeometric2F1[(5 - m)/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m*(1 +
I*Tan[c + d*x])^((1 - m)/2))/(a*d*m*Sqrt[a + I*a*Tan[c + d*x]])

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]

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 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]))

Rubi steps

\begin{align*} \int \frac{(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^{3/2}} \, 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))^{-\frac{3}{2}+\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)^{-\frac{5}{2}+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{5}{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 )^{\frac{1}{2}-\frac{m}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{5}{2}+\frac{m}{2}} (a-i a x)^{-1+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{i 2^{\frac{1}{2} (-3+m)} \, _2F_1\left (\frac{5-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))^{\frac{1-m}{2}}}{a d m \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.58322, size = 178, normalized size = 1.63 \[ -\frac{i 2^{m-\frac{3}{2}} \sqrt{e^{i d x}} e^{-2 i (c+2 d x)} \left (1+e^{2 i (c+d x)}\right )^3 \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{1}{2}} (\cos (d x)+i \sin (d x))^{3/2} \sec ^{\frac{3}{2}-m}(c+d x) \text{Hypergeometric2F1}\left (1,1-\frac{m}{2},\frac{m-1}{2},-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m}{d (m-3) (a+i a \tan (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-I)*2^(-3/2 + m)*Sqrt[E^(I*d*x)]*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1/2 + m)*(1 + E^((2*I)*(c + d*
x)))^3*Hypergeometric2F1[1, 1 - m/2, (-1 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(3/2 - m)*(e*Sec[c + d*x])
^m*(Cos[d*x] + I*Sin[d*x])^(3/2))/(d*E^((2*I)*(c + 2*d*x))*(-3 + m)*(a + I*a*Tan[c + d*x])^(3/2))

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Maple [F]  time = 0.337, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sec \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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))^(3/2),x)

[Out]

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^m/(I*a*tan(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2} \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} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{4 \, a^{2}}, x\right ) \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

integral(1/4*sqrt(2)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(4
*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1)*e^(-3*I*d*x - 3*I*c)/a^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec{\left (c + d x \right )}\right )^{m}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

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))**(3/2),x)

[Out]

Integral((e*sec(c + d*x))**m/(a*(I*tan(c + d*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^m/(I*a*tan(d*x + c) + a)^(3/2), x)