3.5.50 \(\int (e \sec (c+d x))^m (a+i a \tan (c+d x))^5 \, dx\) [450]

Optimal. Leaf size=86 \[ \frac {i 2^{5+\frac {m}{2}} a^5 \, _2F_1\left (-4-\frac {m}{2},\frac {m}{2};\frac {2+m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{d m} \]

[Out]

I*2^(5+1/2*m)*a^5*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/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 a^5 2^{\frac {m}{2}+5} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \, _2F_1\left (-\frac {m}{2}-4,\frac {m}{2};\frac {m+2}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(I*2^(5 + m/2)*a^5*Hypergeometric2F1[-4 - m/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m)/(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 (e \sec (c+d x))^m (a+i a \tan (c+d x))^5 \, 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))^{5+\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}} a^6 (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 )}{d}\\ &=\frac {i 2^{5+\frac {m}{2}} a^5 \, _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}}{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. \(1214\) vs. \(2(86)=172\).
time = 12.26, size = 1214, normalized size = 14.12 \begin {gather*} -\frac {i 2^{5+m} \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 ) \, _2F_1\left (1,1-\frac {m}{2};\frac {2+m}{2};-e^{2 i (c+d x)}\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (e^{3 i c}+e^{5 i c}\right ) m (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{5+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 m x} (2+m) \, _2F_1\left (\frac {m}{2},1+m;\frac {2+m}{2};-e^{2 i (c+d x)}\right )-e^{i d (2+m) x} m \, _2F_1\left (1+m,\frac {2+m}{2};\frac {4+m}{2};-e^{2 i (c+d x)}\right )\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (1+e^{2 i c}\right ) m (2+m) (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{5+m} e^{-3 i c+2 i d x} \left (1+4 e^{2 i c}\right ) \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 \, _2F_1\left (\frac {2+m}{2},2+m;\frac {4+m}{2};-e^{2 i (c+d x)}\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (1+e^{2 i c}\right ) (2+m) (\cos (d x)+i \sin (d x))^5}-\frac {3 i 2^{5+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} (4+m) \, _2F_1\left (\frac {2+m}{2},3+m;\frac {4+m}{2};-e^{2 i (c+d x)}\right )-e^{i d (4+m) x} (2+m) \, _2F_1\left (3+m,\frac {4+m}{2};\frac {6+m}{2};-e^{2 i (c+d x)}\right )\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (1+e^{2 i c}\right ) (2+m) (4+m) (\cos (d x)+i \sin (d x))^5}-\frac {i 2^{5+m} e^{-i (c-4 d x)} \left (2+3 e^{2 i c}\right ) \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 \, _2F_1\left (\frac {4+m}{2},4+m;\frac {6+m}{2};-e^{2 i (c+d x)}\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (1+e^{2 i c}\right ) (4+m) (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{5+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 (4+m) x} (6+m) \, _2F_1\left (\frac {4+m}{2},5+m;\frac {6+m}{2};-e^{2 i (c+d x)}\right )-e^{i d (6+m) x} (4+m) \, _2F_1\left (5+m,\frac {6+m}{2};\frac {8+m}{2};-e^{2 i (c+d x)}\right )\right ) \sec ^{-5-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^5}{d \left (1+e^{2 i c}\right ) (4+m) (6+m) (\cos (d x)+i \sin (d x))^5} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-I)*2^(5 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(1 + E^((2*I)*(c + d*x)))*Hypergeometric2F1[1, 1
- m/2, (2 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-5 - m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d
*(E^((3*I)*c) + E^((5*I)*c))*m*(Cos[d*x] + I*Sin[d*x])^5) + (I*2^(5 + 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*m*x)*(2 + m)*Hypergeometric2F1[m/2, 1 + m, (2 + m)/2, -E^((2*I)*(
c + d*x))] - E^(I*d*(2 + m)*x)*m*Hypergeometric2F1[1 + m, (2 + m)/2, (4 + m)/2, -E^((2*I)*(c + d*x))])*Sec[c +
d*x]^(-5 - m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d*E^(I*(3*c + d*m*x))*(1 + E^((2*I)*c))*m*(2 + m)
*(Cos[d*x] + I*Sin[d*x])^5) + (I*2^(5 + m)*E^((-3*I)*c + (2*I)*d*x)*(1 + 4*E^((2*I)*c))*(E^(I*(c + d*x))/(1 +
E^((2*I)*(c + d*x))))^m*(1 + E^((2*I)*(c + d*x)))^m*Hypergeometric2F1[(2 + m)/2, 2 + m, (4 + m)/2, -E^((2*I)*(
c + d*x))]*Sec[c + d*x]^(-5 - m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d*(1 + E^((2*I)*c))*(2 + m)*(Co
s[d*x] + I*Sin[d*x])^5) - ((3*I)*2^(5 + 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)*(4 + m)*Hypergeometric2F1[(2 + m)/2, 3 + m, (4 + m)/2, -E^((2*I)*(c + d*x))] - E^(I*
d*(4 + m)*x)*(2 + m)*Hypergeometric2F1[3 + m, (4 + m)/2, (6 + m)/2, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^(-5 -
m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d*E^(I*(c + d*m*x))*(1 + E^((2*I)*c))*(2 + m)*(4 + m)*(Cos[d*
x] + I*Sin[d*x])^5) - (I*2^(5 + m)*(2 + 3*E^((2*I)*c))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(1 + E^((
2*I)*(c + d*x)))^m*Hypergeometric2F1[(4 + m)/2, 4 + m, (6 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-5 - m)*
(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d*E^(I*(c - 4*d*x))*(1 + E^((2*I)*c))*(4 + m)*(Cos[d*x] + I*Sin[
d*x])^5) + (I*2^(5 + m)*E^(I*(c - d*m*x))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(1 + E^((2*I)*(c + d*x
)))^m*(E^(I*d*(4 + m)*x)*(6 + m)*Hypergeometric2F1[(4 + m)/2, 5 + m, (6 + m)/2, -E^((2*I)*(c + d*x))] - E^(I*d
*(6 + m)*x)*(4 + m)*Hypergeometric2F1[5 + m, (6 + m)/2, (8 + m)/2, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^(-5 - m
)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^5)/(d*(1 + E^((2*I)*c))*(4 + m)*(6 + m)*(Cos[d*x] + I*Sin[d*x])^5)

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Maple [F]
time = 0.46, size = 0, normalized size = 0.00 \[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{5}\, dx\]

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))^5,x)

[Out]

int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^5,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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))^5,x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^5*(e*sec(d*x + c))^m, x)

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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))^5,x, algorithm="fricas")

[Out]

integral(32*a^5*(2*e^(I*d*x + I*c + 1)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(10*I*d*x + 10*I*c)/(e^(10*I*d*x + 10*I*
c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a^{5} \left (\int \left (- i \left (e \sec {\left (c + d x \right )}\right )^{m}\right )\, dx + \int 5 \left (e \sec {\left (c + d x \right )}\right )^{m} \tan {\left (c + d x \right )}\, dx + \int \left (- 10 \left (e \sec {\left (c + d x \right )}\right )^{m} \tan ^{3}{\left (c + d x \right )}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{m} \tan ^{5}{\left (c + d x \right )}\, dx + \int 10 i \left (e \sec {\left (c + d x \right )}\right )^{m} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 5 i \left (e \sec {\left (c + d x \right )}\right )^{m} \tan ^{4}{\left (c + d x \right )}\right )\, dx\right ) \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))**5,x)

[Out]

I*a**5*(Integral(-I*(e*sec(c + d*x))**m, x) + Integral(5*(e*sec(c + d*x))**m*tan(c + d*x), x) + Integral(-10*(
e*sec(c + d*x))**m*tan(c + d*x)**3, x) + Integral((e*sec(c + d*x))**m*tan(c + d*x)**5, x) + Integral(10*I*(e*s
ec(c + d*x))**m*tan(c + d*x)**2, x) + Integral(-5*I*(e*sec(c + d*x))**m*tan(c + d*x)**4, x))

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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))^5,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^5*(e*sec(d*x + c))^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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 )}^5 \,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)^5,x)

[Out]

int((e/cos(c + d*x))^m*(a + a*tan(c + d*x)*1i)^5, x)

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