\(\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx\) [217]
Optimal result
Integrand size = 28, antiderivative size = 156 \[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}
\]
[Out]
-42/5*a^4*(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/e^2/cos(d*x+
c)^(1/2)/(e*sec(d*x+c))^(1/2)+42/5*a^4*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d/e^3-4/5*I*a*(a+I*a*tan(d*x+c))^3/d/(e
*sec(d*x+c))^(5/2)+28/5*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.19 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3577, 3853, 3856, 2719}
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=\frac {42 a^4 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d e^3}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}
\]
[In]
Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(5/2),x]
[Out]
(-42*a^4*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (42*a^4*Sqrt[e*Sec[c +
d*x]]*Sin[c + d*x])/(5*d*e^3) - (((4*I)/5)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(5/2)) + (((28*I)/
5)*(a^4 + I*a^4*Tan[c + d*x]))/(d*e^2*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*}
\text {integral}& = -\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (7 a^2\right ) \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = -\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}+\frac {\left (21 a^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 e^4} \\ & = \frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {\left (21 a^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2} \\ & = \frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {\left (21 a^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = -\frac {42 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {42 a^4 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d e^3}-\frac {4 i a (a+i a \tan (c+d x))^3}{5 d (e \sec (c+d x))^{5/2}}+\frac {28 i \left (a^4+i a^4 \tan (c+d x)\right )}{5 d e^2 \sqrt {e \sec (c+d x)}} \\
\end{align*}
Mathematica [C] (verified)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.01 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {4 i a^4 e^{2 i (c+d x)} \left (7+2 e^{2 i (c+d x)}-7 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}}
\]
[In]
Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(5/2),x]
[Out]
(((-4*I)/5)*a^4*E^((2*I)*(c + d*x))*(7 + 2*E^((2*I)*(c + d*x)) - 7*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometri
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*e^2*(1 + E^((2*I)*(c + d*x)))*Sqrt[e*Sec[c + d*x]])
Maple [B] (verified)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf
count of optimal. 1720 vs. \(2 (160 ) = 320\).
Time = 22.49 (sec) , antiderivative size = 1721, normalized size of antiderivative =
11.03
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| method | result | size |
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| parts |
\(\text {Expression too large to display}\) |
\(1721\) |
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\(\text {Expression too large to display}\) |
\(1949\) |
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[In]
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
-2/5*a^4/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)/e^2*(3*I*cos(d*x+c)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(
cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*cos(d*x+c)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+6*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(c
sc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)-6*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+3*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*
x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)-3*I*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-cos(d*x+c)^2*sin(d*x+c)-sin(d*x+c)*cos(d*x+c)-3*sin(d*x+c))+2/
5*a^4/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)/e^2*(-12*I*cos(d*x+c)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(c
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+12*I*cos(d*x+c)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-24*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+24*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x
+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)^2*sin(d*x+c)-12*I*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+12*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)*cos(d*x+c)-7*sin(d*x+c)
+5*tan(d*x+c))+2/5*I*a^4/d/(e*sec(d*x+c))^(1/2)/e^2*(5*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*ln((2*cos(d*x+c)*(
-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))-5*(-c
os(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*ln(2*(2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(
d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))-4*cos(d*x+c)^2+5*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1
/2)*ln((2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)
/(cos(d*x+c)+1))-5*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*ln(2*(2*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-cos(d*x+c)+1)/(cos(d*x+c)+1))+20)-8/5*I*a^4/d/(e*sec(d*x+c
))^(5/2)+12/5*a^4/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)/e^2*(-2*I*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(c
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2*I*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),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*I*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)
*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+4*I*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos
(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)^2*sin(d*x+c)-2*I*sec(d*x+c)*(1/(cos(d*x+c)+1))^
(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+2*I*sec(d*x+c)*(cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)*cos(d*x+c)-2*sin(
d*x+c))
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.60
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (21 i \, \sqrt {2} a^{4} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (2 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a^{4} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )}}{5 \, d e^{3}}
\]
[In]
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
-2/5*(21*I*sqrt(2)*a^4*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(2)*(
2*I*a^4*e^(3*I*d*x + 3*I*c) + 7*I*a^4*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*
c))/(d*e^3)
Sympy [F]
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=a^{4} \left (\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {6 \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {4 i \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {4 i \tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right )
\]
[In]
integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(5/2),x)
[Out]
a**4*(Integral((e*sec(c + d*x))**(-5/2), x) + Integral(-6*tan(c + d*x)**2/(e*sec(c + d*x))**(5/2), x) + Integr
al(tan(c + d*x)**4/(e*sec(c + d*x))**(5/2), x) + Integral(4*I*tan(c + d*x)/(e*sec(c + d*x))**(5/2), x) + Integ
ral(-4*I*tan(c + d*x)**3/(e*sec(c + d*x))**(5/2), x))
Maxima [F]
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(5/2), x)
Giac [F]
\[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(5/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x
\]
[In]
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(5/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(5/2), x)