3.450 \(\int (e \sec (c+d x))^m (a+i a \tan (c+d x))^5 \, dx\)

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

[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 = 0.15, 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} \[ \frac {i a^5 2^{\frac {m}{2}+5} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \text {Hypergeometric2F1}\left (-\frac {m}{2}-4,\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \]

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 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 (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 ) \operatorname {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 ) \operatorname {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]  time = 12.92, size = 1165, normalized size = 13.55 \[ -\frac {i 2^{m+5} 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 \, _2F_1\left (1,-\frac {m}{2}-1;\frac {m+6}{2};-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^3 (m+4) (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{m+5} e^{i (c-d m x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (e^{i d (m+4) x} (m+6) \, _2F_1\left (1,-\frac {m}{2}-2;\frac {m+6}{2};-e^{2 i (c+d x)}\right )-e^{i d (m+6) x} (m+4) \, _2F_1\left (1,-\frac {m}{2}-1;\frac {m+8}{2};-e^{2 i (c+d x)}\right )\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^4 (m+4) (m+6) (\cos (d x)+i \sin (d x))^5}-\frac {i 2^{m+5} \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 {m+2}{2};-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (e^{3 i c}+e^{5 i c}\right ) m (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{m+5} e^{-i (3 c+d m x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (e^{i d m x} (m+2) \, _2F_1\left (1,-\frac {m}{2};\frac {m+2}{2};-e^{2 i (c+d x)}\right )-e^{i d (m+2) x} m \, _2F_1\left (1,1-\frac {m}{2};\frac {m+4}{2};-e^{2 i (c+d x)}\right )\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (1+e^{2 i c}\right ) m (m+2) (\cos (d x)+i \sin (d x))^5}+\frac {i 2^{m+5} e^{i (d x-4 c)} \left (1+4 e^{2 i c}\right ) \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+1} \, _2F_1\left (1,-\frac {m}{2};\frac {m+4}{2};-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (1+e^{2 i c}\right ) (m+2) (\cos (d x)+i \sin (d x))^5}-\frac {3 i 2^{m+5} e^{-i (c+d m x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (e^{i d (m+2) x} (m+4) \, _2F_1\left (1,-\frac {m}{2}-1;\frac {m+4}{2};-e^{2 i (c+d x)}\right )-e^{i d (m+4) x} (m+2) \, _2F_1\left (1,-\frac {m}{2};\frac {m+6}{2};-e^{2 i (c+d x)}\right )\right ) (e \sec (c+d x))^m (i \tan (c+d x) a+a)^5 \sec ^{-m-5}(c+d x)}{d \left (1+e^{2 i c}\right ) \left (1+e^{2 i (c+d x)}\right )^2 (m+2) (m+4) (\cos (d x)+i \sin (d x))^5} \]

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)*(2 + 3*E^((2*I)*c))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*Hypergeometric2F1[1, -1 - 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 - 4*d*x))*(1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^3*(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*(E^(I*d*(4 + m)*x)*(6 + m)*Hypergeometric2F1[1
, -2 - m/2, (6 + m)/2, -E^((2*I)*(c + d*x))] - E^(I*d*(6 + m)*x)*(4 + m)*Hypergeometric2F1[1, -1 - 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))*(1 + E^((2*I)*(c + d*x)))^4*(4 + m)*(6 + 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)))*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*(Co
s[d*x] + I*Sin[d*x])^5) + (I*2^(5 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(-(E^(I*d*(2 + m)*x)*m*Hy
pergeometric2F1[1, 1 - m/2, (4 + m)/2, -E^((2*I)*(c + d*x))]) + E^(I*d*m*x)*(2 + m)*Hypergeometric2F1[1, -1/2*
m, (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^(
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^(I*(-4*c + d*x))*(1 +
4*E^((2*I)*c))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1 + m)*Hypergeometric2F1[1, -1/2*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)*(Cos[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*(E^(I*d*(2 +
m)*x)*(4 + m)*Hypergeometric2F1[1, -1 - m/2, (4 + m)/2, -E^((2*I)*(c + d*x))] - E^(I*d*(4 + m)*x)*(2 + m)*Hyp
ergeometric2F1[1, -1/2*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 + d*m*x))*(1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^2*(2 + m)*(4 + m)*(Cos[d*x]
+ I*Sin[d*x])^5)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {32 \, a^{5} \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} e^{\left (10 i \, d x + 10 i \, c\right )}}{e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

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))^5,x, algorithm="fricas")

[Out]

integral(32*a^5*(2*e*e^(I*d*x + I*c)/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \]

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))^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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maple [F]  time = 1.21, 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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \]

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))^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^5 \,d x \]

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

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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