3.681 \(\int \frac{(e \cos (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=175 \[ -\frac{12 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{32 i \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac{16 i \sec ^2(c+d x) (e \cos (c+d x))^{5/2}}{35 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(((2*I)/7)*(e*Cos[c + d*x])^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*
d) - (((32*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)

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Rubi [A]  time = 0.378285, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3515, 3502, 3497, 3488} \[ -\frac{12 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{32 i \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac{16 i \sec ^2(c+d x) (e \cos (c+d x))^{5/2}}{35 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((2*I)/7)*(e*Cos[c + d*x])^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*
d) - (((32*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)

Rule 3515

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 3502

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]

Rule 3497

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] + Dist[(a*(m + n))/(m*d^2), 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] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3488

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]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\left ((e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac{1}{(e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (6 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}-\frac{12 i (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}{35 a d}+\frac{\left (24 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac{1}{\sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{35 e^2}\\ &=\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{12 i (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}{35 a d}+\frac{\left (16 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}} \, dx}{35 a e^2}\\ &=\frac{2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt{a+i a \tan (c+d x)}}+\frac{16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{12 i (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}{35 a d}-\frac{32 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 a d}\\ \end{align*}

Mathematica [A]  time = 0.530145, size = 80, normalized size = 0.46 \[ -\frac{i e^3 (70 i \sin (c+d x)+6 i \sin (3 (c+d x))+35 \cos (c+d x)+\cos (3 (c+d x)))}{70 d \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-I/70)*e^3*(35*Cos[c + d*x] + Cos[3*(c + d*x)] + (70*I)*Sin[c + d*x] + (6*I)*Sin[3*(c + d*x)]))/(d*Sqrt[e*Co
s[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.336, size = 110, normalized size = 0.6 \begin{align*}{\frac{10\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+10\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -32\,i}{35\,ad \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

2/35/d/a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e*cos(d*x+c))^(5/2)*(5*I*cos(d*x+c)^4+5*cos(d*x+c)^3*
sin(d*x+c)+2*I*cos(d*x+c)^2+8*cos(d*x+c)*sin(d*x+c)-16*I)/cos(d*x+c)^2

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Maxima [A]  time = 3.24811, size = 273, normalized size = 1.56 \begin{align*} \frac{{\left (5 i \, e^{2} \cos \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) - 7 i \, e^{2} \cos \left (\frac{5}{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 ) + 35 i \, e^{2} \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 ) - 105 i \, e^{2} \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 ) + 5 \, e^{2} \sin \left (\frac{7}{2} \, d x + \frac{7}{2} \, c\right ) + 7 \, e^{2} \sin \left (\frac{5}{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 ) + 35 \, e^{2} \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 ) + 105 \, e^{2} \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 )} \sqrt{e}}{140 \, \sqrt{a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/140*(5*I*e^2*cos(7/2*d*x + 7/2*c) - 7*I*e^2*cos(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3
5*I*e^2*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 105*I*e^2*cos(1/7*arctan2(sin(7/2*d*x +
7/2*c), cos(7/2*d*x + 7/2*c))) + 5*e^2*sin(7/2*d*x + 7/2*c) + 7*e^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos
(7/2*d*x + 7/2*c))) + 35*e^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*e^2*sin(1/7*ar
ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*sqrt(e)/(sqrt(a)*d)

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Fricas [A]  time = 2.13451, size = 305, normalized size = 1.74 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{1}{2}}{\left (-7 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 105 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 35 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac{7}{2} i \, d x - \frac{7}{2} i \, c\right )}}{140 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/140*sqrt(2)*sqrt(1/2)*(-7*I*e^2*e^(6*I*d*x + 6*I*c) - 105*I*e^2*e^(4*I*d*x + 4*I*c) + 35*I*e^2*e^(2*I*d*x +
2*I*c) + 5*I*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-7/2*I*d*x - 7/2*I*c)/(
a*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/2)/sqrt(I*a*tan(d*x + c) + a), x)