3.192 \(\int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac {10 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}+\frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}}-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}} \]

[Out]

-2/7*I*a/d/(e*sec(d*x+c))^(7/2)+2/7*a*sin(d*x+c)/d/e/(e*sec(d*x+c))^(5/2)+10/21*a*sin(d*x+c)/d/e^3/(e*sec(d*x+
c))^(1/2)+10/21*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*
x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/d/e^4

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Rubi [A]  time = 0.08, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3486, 3769, 3771, 2641} \[ \frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}}+\frac {10 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])/(e*Sec[c + d*x])^(7/2),x]

[Out]

(((-2*I)/7)*a)/(d*(e*Sec[c + d*x])^(7/2)) + (10*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(21*d*e^4) + (2*a*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (10*a*Sin[c + d*x])/(21*d*e^3*Sqrt[e*S
ec[c + d*x]])

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+a \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx\\ &=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac {(5 a) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2}\\ &=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}}+\frac {(5 a) \int \sqrt {e \sec (c+d x)} \, dx}{21 e^4}\\ &=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}}+\frac {\left (5 a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 e^4}\\ &=-\frac {2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac {10 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 d e^4}+\frac {2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac {10 a \sin (c+d x)}{21 d e^3 \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 121, normalized size = 0.97 \[ \frac {a \sqrt {e \sec (c+d x)} (\cos (c+d x)+i \sin (c+d x)) \left (5 \sin (c+d x)+5 \sin (3 (c+d x))-14 i \cos (c+d x)+2 i \cos (3 (c+d x))+20 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)-i \sin (c+d x))\right )}{42 d e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])/(e*Sec[c + d*x])^(7/2),x]

[Out]

(a*Sqrt[e*Sec[c + d*x]]*(Cos[c + d*x] + I*Sin[c + d*x])*((-14*I)*Cos[c + d*x] + (2*I)*Cos[3*(c + d*x)] + 20*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] - I*Sin[c + d*x]) + 5*Sin[c + d*x] + 5*Sin[3*(c + d*x
)]))/(42*d*e^4)

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \frac {{\left (84 \, d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm integral}\left (-\frac {5 i \, \sqrt {2} a \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{21 \, d e^{4}}, x\right ) + \sqrt {2} {\left (-3 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 19 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{84 \, d e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/84*(84*d*e^4*e^(2*I*d*x + 2*I*c)*integral(-5/21*I*sqrt(2)*a*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x
- 1/2*I*c)/(d*e^4), x) + sqrt(2)*(-3*I*a*e^(6*I*d*x + 6*I*c) - 19*I*a*e^(4*I*d*x + 4*I*c) - 9*I*a*e^(2*I*d*x +
2*I*c) + 7*I*a)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(d*e^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)/(e*sec(d*x + c))^(7/2), x)

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maple [A]  time = 0.84, size = 187, normalized size = 1.50 \[ \frac {2 a \left (-3 i \left (\cos ^{4}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )+3 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{21 d \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(7/2),x)

[Out]

2/21*a/d*(-3*I*cos(d*x+c)^4+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos
(d*x+c))/sin(d*x+c),I)*cos(d*x+c)+3*cos(d*x+c)^3*sin(d*x+c)+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*cos(d*x+c)*sin(d*x+c))/(e/cos(d*x+c))^(7/2)/cos(d*x+c
)^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)/(e*sec(d*x + c))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/(e/cos(c + d*x))^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)/(e/cos(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\right )\, dx + \int \frac {\tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))**(7/2),x)

[Out]

I*a*(Integral(-I/(e*sec(c + d*x))**(7/2), x) + Integral(tan(c + d*x)/(e*sec(c + d*x))**(7/2), x))

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