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

Optimal. Leaf size=93 \[ -\frac {i 2^{n+\frac {1}{4}} (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^n \, _2F_1\left (-\frac {3}{4},\frac {7}{4}-n;\frac {1}{4};\frac {1}{2} (1-i \tan (c+d x))\right )}{3 d (e \sec (c+d x))^{3/2}} \]

[Out]

-1/3*I*2^(1/4+n)*hypergeom([-3/4, 7/4-n],[1/4],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+c))^(3/4-n)*(a+I*a*tan(d*x+c
))^n/d/(e*sec(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 93, normalized size of antiderivative = 1.00, 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} \[ -\frac {i 2^{n+\frac {1}{4}} (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^n \text {Hypergeometric2F1}\left (-\frac {3}{4},\frac {7}{4}-n,\frac {1}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{3 d (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/3)*2^(1/4 + n)*Hypergeometric2F1[-3/4, 7/4 - n, 1/4, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(3/4 -
n)*(a + I*a*Tan[c + d*x])^n)/(d*(e*Sec[c + d*x])^(3/2))

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 \frac {(a+i a \tan (c+d x))^n}{(e \sec (c+d x))^{3/2}} \, dx &=\frac {\left ((a-i a \tan (c+d x))^{3/4} (a+i a \tan (c+d x))^{3/4}\right ) \int \frac {(a+i a \tan (c+d x))^{-\frac {3}{4}+n}}{(a-i a \tan (c+d x))^{3/4}} \, dx}{(e \sec (c+d x))^{3/2}}\\ &=\frac {\left (a^2 (a-i a \tan (c+d x))^{3/4} (a+i a \tan (c+d x))^{3/4}\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{-\frac {7}{4}+n}}{(a-i a x)^{7/4}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{3/2}}\\ &=\frac {\left (2^{-\frac {7}{4}+n} a (a-i a \tan (c+d x))^{3/4} (a+i a \tan (c+d x))^n \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {3}{4}-n}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {7}{4}+n}}{(a-i a x)^{7/4}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{3/2}}\\ &=-\frac {i 2^{\frac {1}{4}+n} \, _2F_1\left (-\frac {3}{4},\frac {7}{4}-n;\frac {1}{4};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {3}{4}-n} (a+i a \tan (c+d x))^n}{3 d (e \sec (c+d x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 10.34, size = 129, normalized size = 1.39 \[ -\frac {i 2^{n-\frac {1}{2}} e^{i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n-\frac {5}{2}} \, _2F_1\left (1,\frac {7}{4};n+\frac {1}{4};-e^{2 i (c+d x)}\right ) \sec ^{\frac {3}{2}-n}(c+d x) (a+i a \tan (c+d x))^n}{d (4 n-3) (e \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(-1/2 + n)*E^(I*(c + d*x))*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(-5/2 + n)*Hypergeometric2F1[1,
7/4, 1/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(3/2 - n)*(a + I*a*Tan[c + d*x])^n)/(d*(-3 + 4*n)*(e*Sec[c +
d*x])^(3/2))

________________________________________________________________________________________

fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{4 \, e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.87, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n}}{\left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(3/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________