3.5.71 \(\int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx\) [471]

Optimal. Leaf size=94 \[ \frac {i 2^{\frac {5}{2}+n} a^2 \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-n;\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d} \]

[Out]

1/5*I*2^(5/2+n)*a^2*hypergeom([5/2, -3/2-n],[7/2],1/2-1/2*I*tan(d*x+c))*sec(d*x+c)^5*(1+I*tan(d*x+c))^(-1/2-n)
*(a+I*a*tan(d*x+c))^(-2+n)/d

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Rubi [A]
time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3586, 3604, 72, 71} \begin {gather*} \frac {i a^2 2^{n+\frac {5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-2} \, _2F_1\left (\frac {5}{2},-n-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*2^(5/2 + n)*a^2*Hypergeometric2F1[5/2, -3/2 - n, 7/2, (1 - I*Tan[c + d*x])/2]*Sec[c + d*x]^5*(1 + I*Tan
[c + d*x])^(-1/2 - n)*(a + I*a*Tan[c + d*x])^(-2 + n))/d

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 \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx &=\frac {\sec ^5(c+d x) \int (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{\frac {5}{2}+n} \, dx}{(a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac {\left (a^2 \sec ^5(c+d x)\right ) \text {Subst}\left (\int (a-i a x)^{3/2} (a+i a x)^{\frac {3}{2}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac {\left (2^{\frac {3}{2}+n} a^3 \sec ^5(c+d x) (a+i a \tan (c+d x))^{-2+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{\frac {3}{2}+n} (a-i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2}}\\ &=\frac {i 2^{\frac {5}{2}+n} a^2 \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-n;\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 14.34, size = 153, normalized size = 1.63 \begin {gather*} -\frac {i 2^{5+n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \, _2F_1\left (\frac {5}{2}+n,5+n;\frac {7}{2}+n;-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (5+2 n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^(5 + n)*E^((5*I)*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(1 + E^((2*I)*
(c + d*x)))^n*Hypergeometric2F1[5/2 + n, 5 + n, 7/2 + n, -E^((2*I)*(c + d*x))]*(a + I*a*Tan[c + d*x])^n)/(d*(5
+ 2*n)*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]
time = 0.34, size = 0, normalized size = 0.00 \[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,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(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^n*sec(d*x + c)^5, 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(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(32*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*e^(5*I*d*x + 5*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*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{5}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c))**n,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*sec(c + d*x)**5, 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(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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