3.225 \(\int \frac {(e \sec (c+d x))^{7/2}}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=101 \[ -\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{a d}-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 101, 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 = {3501, 3768, 3771, 2639} \[ -\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{a d}-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2639

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

Rule 3501

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

Rule 3768

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 3771

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 {(e \sec (c+d x))^{7/2}}{a+i a \tan (c+d x)} \, dx &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {e^2 \int (e \sec (c+d x))^{3/2} \, dx}{a}\\ &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}-\frac {e^4 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{a}\\ &=-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}-\frac {e^4 \int \sqrt {\cos (c+d x)} \, dx}{a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {2 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {2 i e^2 (e \sec (c+d x))^{3/2}}{3 a d}+\frac {2 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a d}\\ \end {align*}

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Mathematica [C]  time = 0.84, size = 102, normalized size = 1.01 \[ \frac {2 i e^3 (\cos (c)+i \sin (c)) (\cos (d x)+i \sin (d x)) \sqrt {e \sec (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+i \tan (c+d x)-4\right )}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/3)*e^3*Sqrt[e*Sec[c + d*x]]*(Cos[c] + I*Sin[c])*(Cos[d*x] + I*Sin[d*x])*(-4 + Sqrt[1 + E^((2*I)*(c + d
*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + I*Tan[c + d*x]))/(a*d)

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-6 i \, e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 10 i \, e^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 3 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} {\rm integral}\left (\frac {i \, \sqrt {2} e^{3} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{a d}, x\right )}{3 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(2)*(-6*I*e^3*e^(3*I*d*x + 3*I*c) - 10*I*e^3*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) + 3*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*integral(I*sqrt(2)*e^3*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1)
)*e^(1/2*I*d*x + 1/2*I*c)/(a*d), x))/(a*d*e^(2*I*d*x + 2*I*c) + a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.09, size = 361, normalized size = 3.57 \[ \frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 \left (\cos ^{2}\left (d x +c \right )\right )-i \sin \left (d x +c \right )+3 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\cos ^{2}\left (d x +c \right )\right )}{3 a d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c)),x)

[Out]

2/3/a/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(3*I*sin(d*x+c)*cos(d*x+c)^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)-3*I*sin(d*x+c)*cos(d*x+c)^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)+3*I*cos(d*x+c)*(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)-3*I*cos(d*x
+c)*(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)-3*cos(d*x+c)^2-I*sin(d*x+c)+3*cos(d*x+c))*(e/cos(d*x+c))^(7/2)*cos(d*x+c)^2/sin(d*x+c)^5

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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