3.5.56 \(\int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx\) [456]

Optimal. Leaf size=86 \[ \frac {i 2^{-3+\frac {m}{2}} \, _2F_1\left (4-\frac {m}{2},\frac {m}{2};\frac {2+m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{a^3 d m} \]

[Out]

I*2^(-3+1/2*m)*hypergeom([1/2*m, 4-1/2*m],[1+1/2*m],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^m/a^3/d/m/((1+I*tan(d
*x+c))^(1/2*m))

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Rubi [A]
time = 0.12, 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}-3} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \, _2F_1\left (4-\frac {m}{2},\frac {m}{2};\frac {m+2}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{a^3 d m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^(-3 + m/2)*Hypergeometric2F1[4 - m/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m)/(a^3*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))^3} \, 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))^{-3+\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)^{-4+\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-4+\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 )^{-4+\frac {m}{2}} (a-i a x)^{-1+\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=\frac {i 2^{-3+\frac {m}{2}} \, _2F_1\left (4-\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^3 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. \(347\) vs. \(2(86)=172\).
time = 10.14, size = 347, normalized size = 4.03 \begin {gather*} -\frac {i 2^{-3+m} e^{-i (3 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 (-6+m) x} m \left (8-6 m+m^2\right ) \, _2F_1\left (\frac {1}{2} (-6+m),m;\frac {1}{2} (-4+m);-e^{2 i (c+d x)}\right )+e^{2 i c} (-6+m) \left (3 e^{i d (-4+m) x} (-2+m) m \, _2F_1\left (\frac {1}{2} (-4+m),m;\frac {1}{2} (-2+m);-e^{2 i (c+d x)}\right )+e^{2 i c} (-4+m) \left (3 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 )\right )\right ) \sec ^{3-m}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))^3}{d (-6+m) (-4+m) (-2+m) m (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^(-3 + 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*(-6 + m)*x)
*m*(8 - 6*m + m^2)*Hypergeometric2F1[(-6 + m)/2, m, (-4 + m)/2, -E^((2*I)*(c + d*x))] + E^((2*I)*c)*(-6 + m)*(
3*E^(I*d*(-4 + m)*x)*(-2 + m)*m*Hypergeometric2F1[(-4 + m)/2, m, (-2 + m)/2, -E^((2*I)*(c + d*x))] + E^((2*I)*
c)*(-4 + m)*(3*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)*Hypergeometric2F1[m/2, m, (2 + m)/2, -E^((2*I)*(c + d*x))])))*Sec[c + d*x]^(3 - m)*(e*Sec[c +
d*x])^m*(Cos[d*x] + I*Sin[d*x])^3)/(d*E^(I*(3*c + d*m*x))*(-6 + m)*(-4 + m)*(-2 + m)*m*(a + I*a*Tan[c + d*x])
^3)

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Maple [F]
time = 0.61, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{\left (a +i a \tan \left (d x +c \right )\right )^{3}}\, dx\]

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))^3,x)

[Out]

int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,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))^3,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))^3,x, algorithm="fricas")

[Out]

integral(1/8*(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1))^m*(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c)
+ 3*e^(2*I*d*x + 2*I*c) + 1)*e^(-6*I*d*x - 6*I*c)/a^3, 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 ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \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))**3,x)

[Out]

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

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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))^3,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^m/(I*a*tan(d*x + c) + a)^3, 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}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,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)^3,x)

[Out]

int((e/cos(c + d*x))^m/(a + a*tan(c + d*x)*1i)^3, x)

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