3.214 \(\int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=183 \[ \frac{78 i a^4 \sqrt{e \sec (c+d x)}}{7 d}+\frac{26 i \left (a^2+i a^2 \tan (c+d x)\right )^2 \sqrt{e \sec (c+d x)}}{35 d}+\frac{78 i \left (a^4+i a^4 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}{35 d}+\frac{78 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d}+\frac{2 i a (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}}{7 d} \]

[Out]

(((78*I)/7)*a^4*Sqrt[e*Sec[c + d*x]])/d + (78*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(7*d) + (((2*I)/7)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3)/d + (((26*I)/35)*Sqrt[e*Sec[c + d*x
]]*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (((78*I)/35)*Sqrt[e*Sec[c + d*x]]*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.201563, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3498, 3486, 3771, 2641} \[ \frac{78 i a^4 \sqrt{e \sec (c+d x)}}{7 d}+\frac{26 i \left (a^2+i a^2 \tan (c+d x)\right )^2 \sqrt{e \sec (c+d x)}}{35 d}+\frac{78 i \left (a^4+i a^4 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}{35 d}+\frac{78 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d}+\frac{2 i a (a+i a \tan (c+d x))^3 \sqrt{e \sec (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((78*I)/7)*a^4*Sqrt[e*Sec[c + d*x]])/d + (78*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(7*d) + (((2*I)/7)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^3)/d + (((26*I)/35)*Sqrt[e*Sec[c + d*x
]]*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (((78*I)/35)*Sqrt[e*Sec[c + d*x]]*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3498

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

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 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]

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]

Rubi steps

\begin{align*} \int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx &=\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{1}{7} (13 a) \int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{26 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac{1}{35} \left (117 a^2\right ) \int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx\\ &=\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{26 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac{78 i \sqrt{e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}+\frac{1}{7} \left (39 a^3\right ) \int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx\\ &=\frac{78 i a^4 \sqrt{e \sec (c+d x)}}{7 d}+\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{26 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac{78 i \sqrt{e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}+\frac{1}{7} \left (39 a^4\right ) \int \sqrt{e \sec (c+d x)} \, dx\\ &=\frac{78 i a^4 \sqrt{e \sec (c+d x)}}{7 d}+\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{26 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac{78 i \sqrt{e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}+\frac{1}{7} \left (39 a^4 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{78 i a^4 \sqrt{e \sec (c+d x)}}{7 d}+\frac{78 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 d}+\frac{2 i a \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^3}{7 d}+\frac{26 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )^2}{35 d}+\frac{78 i \sqrt{e \sec (c+d x)} \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}\\ \end{align*}

Mathematica [A]  time = 1.46473, size = 101, normalized size = 0.55 \[ \frac{a^4 \sec ^4(c+d x) \sqrt{e \sec (c+d x)} \left (-150 \sin (2 (c+d x))-85 \sin (4 (c+d x))+1008 i \cos (2 (c+d x))+280 i \cos (4 (c+d x))+1560 \cos ^{\frac{9}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+728 i\right )}{140 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*Sec[c + d*x]^4*Sqrt[e*Sec[c + d*x]]*(728*I + (1008*I)*Cos[2*(c + d*x)] + (280*I)*Cos[4*(c + d*x)] + 1560*
Cos[c + d*x]^(9/2)*EllipticF[(c + d*x)/2, 2] - 150*Sin[2*(c + d*x)] - 85*Sin[4*(c + d*x)]))/(140*d)

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Maple [A]  time = 0.318, size = 230, normalized size = 1.3 \begin{align*}{\frac{2\,{a}^{4} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{35\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( 195\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{4}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +195\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+280\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-85\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -28\,i\cos \left ( dx+c \right ) +5\,\sin \left ( dx+c \right ) \right ) \sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x)

[Out]

2/35*a^4/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(195*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*cos(d*x+c)^4*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+195*I*cos(d*x+c)^3*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c
),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+280*I*cos(d*x+c)^3-85*cos(d*x+c)^2*sin(d*x+c)-
28*I*cos(d*x+c)+5*sin(d*x+c))*(e/cos(d*x+c))^(1/2)/sin(d*x+c)^4/cos(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (730 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 1586 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 1326 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 390 i \, a^{4}\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 )} + 35 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}{\rm integral}\left (-\frac{39 i \, \sqrt{2} a^{4} \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 )}}{7 \, d}, x\right )}{35 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/35*(sqrt(2)*(730*I*a^4*e^(6*I*d*x + 6*I*c) + 1586*I*a^4*e^(4*I*d*x + 4*I*c) + 1326*I*a^4*e^(2*I*d*x + 2*I*c)
+ 390*I*a^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 35*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4
*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*integral(-39/7*I*sqrt(2)*a^4*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*
e^(-1/2*I*d*x - 1/2*I*c)/d, x))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int \sqrt{e \sec{\left (c + d x \right )}}\, dx + \int - 6 \sqrt{e \sec{\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sqrt{e \sec{\left (c + d x \right )}} \tan ^{4}{\left (c + d x \right )}\, dx + \int 4 i \sqrt{e \sec{\left (c + d x \right )}} \tan{\left (c + d x \right )}\, dx + \int - 4 i \sqrt{e \sec{\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**4,x)

[Out]

a**4*(Integral(sqrt(e*sec(c + d*x)), x) + Integral(-6*sqrt(e*sec(c + d*x))*tan(c + d*x)**2, x) + Integral(sqrt
(e*sec(c + d*x))*tan(c + d*x)**4, x) + Integral(4*I*sqrt(e*sec(c + d*x))*tan(c + d*x), x) + Integral(-4*I*sqrt
(e*sec(c + d*x))*tan(c + d*x)**3, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^4, x)