3.5.27 \(\int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx\) [427]

Optimal. Leaf size=121 \[ \frac {2 i}{7 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {8 i}{21 a d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{21 a^2 d \sqrt {e \sec (c+d x)}} \]

[Out]

8/21*I/a/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-16/21*I*(a+I*a*tan(d*x+c))^(1/2)/a^2/d/(e*sec(d*x+c))
^(1/2)+2/7*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.15, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3583, 3569} \begin {gather*} -\frac {16 i \sqrt {a+i a \tan (c+d x)}}{21 a^2 d \sqrt {e \sec (c+d x)}}+\frac {8 i}{21 a d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((2*I)/7)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + ((8*I)/21)/(a*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a
+ I*a*Tan[c + d*x]]) - (((16*I)/21)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d*Sqrt[e*Sec[c + d*x]])

Rule 3569

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/(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), 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] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {2 i}{7 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {4 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{7 a}\\ &=\frac {2 i}{7 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {8 i}{21 a d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{21 a^2}\\ &=\frac {2 i}{7 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {8 i}{21 a d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{21 a^2 d \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 83, normalized size = 0.69 \begin {gather*} -\frac {\sec ^2(c+d x) (-7+9 \cos (2 (c+d x))+12 i \sin (2 (c+d x)))}{21 a d \sqrt {e \sec (c+d x)} (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

-1/21*(Sec[c + d*x]^2*(-7 + 9*Cos[2*(c + d*x)] + (12*I)*Sin[2*(c + d*x)]))/(a*d*Sqrt[e*Sec[c + d*x]]*(-I + Tan
[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 1.17, size = 106, normalized size = 0.88

method	result	size
default	\(-\frac {2 \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (9 i \left (\cos ^{2}\left (d x +c \right )\right )-12 \sin \left (d x +c \right ) \cos \left (d x +c \right )-8 i\right )}{21 d \left (2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-1\right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, a^{2}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/21/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(9*I*cos(d*x+c)^2-12*sin(d*x+c)*cos(d*x+c)-8*I)/(2*I*co
s(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)/(e/cos(d*x+c))^(1/2)/a^2

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Maxima [A]
time = 0.56, size = 129, normalized size = 1.07 \begin {gather*} \frac {{\left (3 i \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 14 i \, \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 21 i \, \cos \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 3 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 14 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 21 \, \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} e^{\left (-\frac {1}{2}\right )}}{42 \, a^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/42*(3*I*cos(7/2*d*x + 7/2*c) + 14*I*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 21*I*cos(
1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3*sin(7/2*d*x + 7/2*c) + 14*sin(3/7*arctan2(sin(7/2
*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 21*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*e^(-1
/2)/(a^(3/2)*d)

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Fricas [A]
time = 0.37, size = 83, normalized size = 0.69 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-21 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c - \frac {1}{2}\right )}}{42 \, a^{2} d \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/42*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-21*I*e^(6*I*d*x + 6*I*c) - 7*I*e^(4*I*d*x + 4*I*c) + 17*I*e^(2*I*d*x
+ 2*I*c) + 3*I)*e^(-7/2*I*d*x - 7/2*I*c - 1/2)/(a^2*d*sqrt(e^(2*I*d*x + 2*I*c) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(1/(sqrt(e*sec(c + d*x))*(I*a*(tan(c + d*x) - I))**(3/2)), 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(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(e^(-1/2)/((I*a*tan(d*x + c) + a)^(3/2)*sqrt(sec(d*x + c))), x)

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Mupad [B]
time = 4.16, size = 104, normalized size = 0.86 \begin {gather*} \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (35\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )-\cos \left (c+d\,x\right )\,7{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{42\,a\,d\,e\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

((e/cos(c + d*x))^(1/2)*(35*sin(c + d*x) - cos(c + d*x)*7i + cos(3*c + 3*d*x)*3i + 3*sin(3*c + 3*d*x)))/(42*a*
d*e*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))

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