3.3.15 \(\int \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx\) [215]

Optimal. Leaf size=178 \[ \frac {154 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {154 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2} \]

[Out]

-154/15*I*a^4*(e*sec(d*x+c))^(3/2)/d/e^2+154/5*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)-154/5*a^4*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d
/e-4*I*a*(a+I*a*tan(d*x+c))^3/d/(e*sec(d*x+c))^(1/2)-22/5*I*(e*sec(d*x+c))^(3/2)*(a^4+I*a^4*tan(d*x+c))/d/e^2

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Rubi [A]
time = 0.13, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3577, 3579, 3567, 3853, 3856, 2719} \begin {gather*} -\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {22 i \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d e^2}-\frac {154 a^4 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d e}+\frac {154 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(154*a^4*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (((154*I)/15)*a^4*(e*Sec[c
+ d*x])^(3/2))/(d*e^2) - (154*a^4*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(5*d*e) - ((4*I)*a*(a + I*a*Tan[c + d*x]
)^3)/(d*Sqrt[e*Sec[c + d*x]]) - (((22*I)/5)*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 \frac {(a+i a \tan (c+d x))^4}{\sqrt {e \sec (c+d x)}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {\left (11 a^2\right ) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx}{e^2}\\ &=-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2}-\frac {\left (77 a^3\right ) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx}{5 e^2}\\ &=-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2}-\frac {\left (77 a^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 e^2}\\ &=-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {154 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2}+\frac {1}{5} \left (77 a^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {154 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2}+\frac {\left (77 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {154 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {154 i a^4 (e \sec (c+d x))^{3/2}}{15 d e^2}-\frac {154 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e}-\frac {4 i a (a+i a \tan (c+d x))^3}{d \sqrt {e \sec (c+d x)}}-\frac {22 i (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 3.99, size = 123, normalized size = 0.69 \begin {gather*} -\frac {2 i a^4 e^{i (c+d x)} \left (-77-176 e^{2 i (c+d x)}-111 e^{4 i (c+d x)}+77 \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sqrt {e \sec (c+d x)}}{15 d e \left (1+e^{2 i (c+d x)}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/15)*a^4*E^(I*(c + d*x))*(-77 - 176*E^((2*I)*(c + d*x)) - 111*E^((4*I)*(c + d*x)) + 77*(1 + E^((2*I)*(
c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sqrt[e*Sec[c + d*x]])/(d*e*(1 + E^((2
*I)*(c + d*x)))^2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1617 vs. \(2 (179 ) = 358\).
time = 0.59, size = 1618, normalized size = 9.09

method	result	size
default	\(\text {Expression too large to display}\)	\(1618\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*a^4/d*(-1+cos(d*x+c))^3*(3*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-120*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/
2)*sin(d*x+c)*cos(d*x+c)^6-360*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^5-380*I*(-cos(d*x+
c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^4-180*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d
*x+c)^3-30*I*cos(d*x+c)^4*ln(-2*(2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+
c)/(1+cos(d*x+c))^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*sin(d*x+c)+30*I*cos(d*x+c)^4*ln(-(2*(-cos(d*x+c)/(1+c
os(d*x+c))^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c
)^2)*sin(d*x+c)-60*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)^2-20*I*(-cos(d*x+c)/(1+cos(d*x
+c))^2)^(3/2)*sin(d*x+c)*cos(d*x+c)-120*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^7+231*I*EllipticF(I*(-
1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1
/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+219*cos(d*x+c)^5*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+231*cos(d*x+c)^4*(
-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-108*cos(d*x+c)^3*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-105*cos(d*x+c)^2*(-c
os(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-129*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-1386*I*EllipticE(I*(-1
+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/
2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+924*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^3*(
-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+9*cos(d*x+c)*(-
cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-924*I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^3*(-c
os(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+231*I*(-cos(d*x+c
)/(1+cos(d*x+c))^2)^(3/2)*(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)*sin(d*x+c)*cos(d*x+c)^2-231*I*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^2-231*I*Ellip
ticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^6*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*
x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+924*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(
d*x+c)^5*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-924*I
*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)*cos(d*x+c)^5*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+
cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+1386*I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+
c)*cos(d*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
))/sin(d*x+c)^7/cos(d*x+c)^3/(e/cos(d*x+c))^(1/2)/(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)

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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((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),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 = 0.10, size = 164, normalized size = 0.92 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-111 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 176 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 77 i \, a^{4} e^{\left (i \, d x + i \, c\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}} + 231 \, {\left (-i \, \sqrt {2} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, \sqrt {2} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {2} a^{4}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (d e^{\frac {1}{2}} + d e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(sqrt(2)*(-111*I*a^4*e^(5*I*d*x + 5*I*c) - 176*I*a^4*e^(3*I*d*x + 3*I*c) - 77*I*a^4*e^(I*d*x + I*c))*e^(
1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1) + 231*(-I*sqrt(2)*a^4*e^(4*I*d*x + 4*I*c) - 2*I*sqrt(2)*a^4
*e^(2*I*d*x + 2*I*c) - I*sqrt(2)*a^4)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))/(d*
e^(1/2) + d*e^(4*I*d*x + 4*I*c + 1/2) + 2*d*e^(2*I*d*x + 2*I*c + 1/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{4} \left (\int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\sqrt {e \sec {\left (c + d x \right )}}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(Integral(1/sqrt(e*sec(c + d*x)), x) + Integral(-6*tan(c + d*x)**2/sqrt(e*sec(c + d*x)), x) + Integral(ta
n(c + d*x)**4/sqrt(e*sec(c + d*x)), x) + Integral(4*I*tan(c + d*x)/sqrt(e*sec(c + d*x)), x) + Integral(-4*I*ta
n(c + d*x)**3/sqrt(e*sec(c + d*x)), 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((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(1/2),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 \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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