3.2.95 \(\int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx\) [195]

Optimal. Leaf size=107 \[ \frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \]

[Out]

6*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)
/(e*sec(d*x+c))^(1/2)-6*a^2*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d/e-4*I*(a^2+I*a^2*tan(d*x+c))/d/(e*sec(d*x+c))^(1
/2)

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Rubi [A]
time = 0.06, antiderivative size = 107, 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 = {3577, 3853, 3856, 2719} \begin {gather*} -\frac {6 a^2 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*a^2*Sqrt[e*Sec[c + d*x]]*Si
n[c + d*x])/(d*e) - ((4*I)*(a^2 + I*a^2*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])

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 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 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))^2}{\sqrt {e \sec (c+d x)}} \, dx &=-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}-\frac {\left (3 a^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{e^2}\\ &=-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\left (3 a^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\frac {\left (3 a^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.01, size = 132, normalized size = 1.23 \begin {gather*} -\frac {2 i \sqrt {2} a^2 e^{2 i (c+d x)} \left (-\sqrt {1+e^{2 i (c+d x)}}+\left (1+e^{2 i (c+d x)}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )}{d \sqrt {\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[2]*a^2*E^((2*I)*(c + d*x))*(-Sqrt[1 + E^((2*I)*(c + d*x))] + (1 + E^((2*I)*(c + d*x)))*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*Sqrt[(e*E^(I*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^(
(2*I)*(c + d*x)))^(3/2))

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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. 1098 vs. \(2 (123 ) = 246\).
time = 0.51, size = 1099, normalized size = 10.27

method	result	size
default	\(\text {Expression too large to display}\)	\(1099\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2/d*(-1+cos(d*x+c))*(12*I*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+6*I*cos(d*x+c)^3*sin(
d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-
cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+12*I*cos(d*x+c)^2*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(
1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*cos(d*x+c)^5*(-c
os(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-12*I*cos(d*x+c)^2*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1
+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*I*cos(d*x+c)^4*si
n(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+I*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)*cos(d*x+c)^2*sin(d*x+c)+6*cos(d
*x+c)^4*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+4*I*cos(d*x+c)*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+6*
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)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*sin(d*x+c)-6*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-4*co
s(d*x+c)^3*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)-I*cos(d*x+c)^2*ln(-(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)-6*I*cos
(d*x+c)^3*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+12*I*cos(d*x+c)^3*sin(d*x+c)*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(
3/2)-8*cos(d*x+c)^2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)+2*(-cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2))/(1+cos(d*x+c)
)^2/cos(d*x+c)/sin(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))^2/(e*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

e^(-1/2)*integrate((I*a*tan(d*x + c) + a)^2/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 = 61, normalized size = 0.57 \begin {gather*} -\frac {2 \, {\left (-3 i \, \sqrt {2} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) - \frac {i \, \sqrt {2} a^{2} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-\frac {1}{2}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-3*I*sqrt(2)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) - I*sqrt(2)*a^2*e^(3/
2*I*d*x + 3/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2)/d

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

[Out]

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

[Out]

integrate((I*a*tan(d*x + c) + a)^2*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 )}^2}{\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)^2/(e/cos(c + d*x))^(1/2),x)

[Out]

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

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