3.3.14 \(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^4 \, dx\) [214]

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

[Out]

78/7*I*a^4*(e*sec(d*x+c))^(1/2)/d+78/7*a^4*(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+2/7*I*a*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^3/d
+26/35*I*(e*sec(d*x+c))^(1/2)*(a^2+I*a^2*tan(d*x+c))^2/d+78/35*I*(e*sec(d*x+c))^(1/2)*(a^4+I*a^4*tan(d*x+c))/d

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Rubi [A]
time = 0.15, antiderivative size = 183, normalized size of antiderivative = 1.00, 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 = {3579, 3567, 3856, 2720} \begin {gather*} \frac {78 i a^4 \sqrt {e \sec (c+d x)}}{7 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 {26 i \left (a^2+i a^2 \tan (c+d x)\right )^2 \sqrt {e \sec (c+d x)}}{35 d}+\frac {2 i a (a+i a \tan (c+d x))^3 \sqrt {e \sec (c+d x)}}{7 d} \end {gather*}

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 2720

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

Rule 3567

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 3579

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 3856

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 \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*}

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Mathematica [A]
time = 1.02, size = 101, normalized size = 0.55 \begin {gather*} \frac {a^4 \sec ^4(c+d x) \sqrt {e \sec (c+d x)} \left (728 i+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 )-150 \sin (2 (c+d x))-85 \sin (4 (c+d x))\right )}{140 d} \end {gather*}

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 = 0.58, size = 230, normalized size = 1.26

method	result	size
default	\(-\frac {2 a^{4} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (-195 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{4}\left (d x +c \right )\right )-195 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{3}\left (d x +c \right )\right )-280 i \left (\cos ^{3}\left (d x +c \right )\right )+85 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+28 i \cos \left (d x +c \right )-5 \sin \left (d x +c \right )\right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}}{35 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{4}}\)	\(230\)

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,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 200, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-195 i \, a^{4} e^{\frac {1}{2}} - 365 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c + \frac {1}{2}\right )} - 793 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} - 663 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 195 \, {\left (i \, \sqrt {2} a^{4} e^{\frac {1}{2}} + i \, \sqrt {2} a^{4} e^{\left (6 i \, d x + 6 i \, c + \frac {1}{2}\right )} + 3 i \, \sqrt {2} a^{4} e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 3 i \, \sqrt {2} a^{4} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\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 {gather*}

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]

-2/35*(sqrt(2)*(-195*I*a^4*e^(1/2) - 365*I*a^4*e^(6*I*d*x + 6*I*c + 1/2) - 793*I*a^4*e^(4*I*d*x + 4*I*c + 1/2)
- 663*I*a^4*e^(2*I*d*x + 2*I*c + 1/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 195*(I*sqrt(2)
*a^4*e^(1/2) + I*sqrt(2)*a^4*e^(6*I*d*x + 6*I*c + 1/2) + 3*I*sqrt(2)*a^4*e^(4*I*d*x + 4*I*c + 1/2) + 3*I*sqrt(
2)*a^4*e^(2*I*d*x + 2*I*c + 1/2))*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{4} \left (\int \sqrt {e \sec {\left (c + d x \right )}}\, dx + \int \left (- 6 \sqrt {e \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\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 \left (- 4 i \sqrt {e \sec {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\right )\, dx\right ) \end {gather*}

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 = 0.00, 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((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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