\(\int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\) [670]

Optimal result

Integrand size = 28, antiderivative size = 122 \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

6*cos(d*x+c)^(7/2)*(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))/a^2/d
/(e*cos(d*x+c))^(7/2)-6*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(7/2)+4*I*cos(d*x+c)^2/d/(e*cos(d*x+c))^(
7/2)/(a^2+I*a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3853, 3856, 2719} \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \sin (c+d x) \cos ^3(c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{7/2}} \]

[In]

Int[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(6*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2])/(a^2*d*(e*Cos[c + d*x])^(7/2)) - (6*Cos[c + d*x]^3*Sin[c + d*
x])/(a^2*d*(e*Cos[c + d*x])^(7/2)) + ((4*I)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[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 3581

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

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[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 {\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}} \\ & = \frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{a^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}} \\ & = -\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 e^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{a^2 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}} \\ & = -\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 \cos ^{\frac {7}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2 (e \cos (c+d x))^{7/2}} \\ & = \frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.39 (sec) , antiderivative size = 1106, normalized size of antiderivative = 9.07 \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\frac {\sqrt {\cos (c+d x)} (\cos (d x)+i \sin (d x))^2 \left (-2 i \cos (c-d x) \sqrt {\cos (c+d x)}+2 \sqrt {\cos (c+d x)} \sin (c-d x)\right )}{d (e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2}+\frac {3 \cos (c) \cos ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-\frac {\cos (d x-\arctan (\cot (c))) \cot (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\sin ^2(d x-\arctan (\cot (c)))\right )}{\sqrt {1+\cot ^2(c)} \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}+\frac {\frac {\cos (d x-\arctan (\cot (c))) \cot (c)}{\sqrt {1+\cot ^2(c)}}+\frac {2 \sqrt {1+\cot ^2(c)} \sin ^2(c) \sin (d x-\arctan (\cot (c)))}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))}}\right )}{d (e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2}+\frac {3 i \cos ^{\frac {3}{2}}(c+d x) \sin (c) (\cos (d x)+i \sin (d x))^2 \left (-\frac {\cos (d x-\arctan (\cot (c))) \cot (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\sin ^2(d x-\arctan (\cot (c)))\right )}{\sqrt {1+\cot ^2(c)} \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}+\frac {\frac {\cos (d x-\arctan (\cot (c))) \cot (c)}{\sqrt {1+\cot ^2(c)}}+\frac {2 \sqrt {1+\cot ^2(c)} \sin ^2(c) \sin (d x-\arctan (\cot (c)))}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))}}\right )}{d (e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2}-\frac {3 i \cos (c) \cos ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2 \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{d (e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2}+\frac {3 \cos ^{\frac {3}{2}}(c+d x) \sin (c) (\cos (d x)+i \sin (d x))^2 \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{d (e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \]

[In]

Integrate[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2*((-2*I)*Cos[c - d*x]*Sqrt[Cos[c + d*x]] + 2*Sqrt[Cos[c + d*x]]*S
in[c - d*x]))/(d*(e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2) + (3*Cos[c]*Cos[c + d*x]^(3/2)*(Cos[d*x] + I
*Sin[d*x])^2*(-((Cos[d*x - ArcTan[Cot[c]]]*Cot[c]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Sin[d*x - ArcTan[Cot[
c]]]^2])/(Sqrt[1 + Cot[c]^2]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - Ar
cTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])) + ((Cos[d*x - ArcTan[Cot[c]]]*Cot[c])/Sqrt[1 + Cot[c]^2]
+ (2*Sqrt[1 + Cot[c]^2]*Sin[c]^2*Sin[d*x - ArcTan[Cot[c]]])/(Cos[c]^2 + Sin[c]^2))/Sqrt[-(Sqrt[1 + Cot[c]^2]*
Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]))/(d*(e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2) + ((3*I)*Cos[c + d*x]
^(3/2)*Sin[c]*(Cos[d*x] + I*Sin[d*x])^2*(-((Cos[d*x - ArcTan[Cot[c]]]*Cot[c]*HypergeometricPFQ[{-1/2, -1/4}, {
3/4}, Sin[d*x - ArcTan[Cot[c]]]^2])/(Sqrt[1 + Cot[c]^2]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Co
t[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])) + ((Cos[d*x - ArcTan[Cot[c]]]
*Cot[c])/Sqrt[1 + Cot[c]^2] + (2*Sqrt[1 + Cot[c]^2]*Sin[c]^2*Sin[d*x - ArcTan[Cot[c]]])/(Cos[c]^2 + Sin[c]^2))
/Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]))/(d*(e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x
])^2) - ((3*I)*Cos[c]*Cos[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Co
s[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos
[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[
d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(
Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(e*Cos[c + d*x])^(7/2)*(a
+ I*a*Tan[c + d*x])^2) + (3*Cos[c + d*x]^(3/2)*Sin[c]*(Cos[d*x] + I*Sin[d*x])^2*((HypergeometricPFQ[{-1/2, -1
/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]
]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c
]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1
+ Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(e*Cos[c +
d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2)

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {2 \left (4 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}-2 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{e^{3} a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(135\)

[In]

int(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-2/e^3/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(4*I*sin(1/2*d*x+1/2*c)^3+2*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^2-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)-2*I*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 3 \, {\left (-i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{a^{2} d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{4}} \]

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*cos(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

int(1/((e*cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

int(1/((e*cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^2), x)