3.673 \(\int (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=179 \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}}{7 d}+\frac{32 i a \sec ^4(c+d x) (e \cos (c+d x))^{7/2}}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}}{35 d}+\frac{12 i a \sec ^2(c+d x) (e \cos (c+d x))^{7/2}}{35 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(((12*I)/35)*a*(e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((32*I)/35)*a*(e*Cos[c
+ d*x])^(7/2)*Sec[c + d*x]^4)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*(e*Cos[c + d*x])^(7/2)*Sqrt[a + I*a
*Tan[c + d*x]])/d - (((16*I)/35)*(e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.382654, antiderivative size = 179, 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, 3497, 3502, 3488} \[ -\frac{2 i \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}}{7 d}+\frac{32 i a \sec ^4(c+d x) (e \cos (c+d x))^{7/2}}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{16 i \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{7/2}}{35 d}+\frac{12 i a \sec ^2(c+d x) (e \cos (c+d x))^{7/2}}{35 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((12*I)/35)*a*(e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((32*I)/35)*a*(e*Cos[c
+ d*x])^(7/2)*Sec[c + d*x]^4)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*(e*Cos[c + d*x])^(7/2)*Sqrt[a + I*a
*Tan[c + d*x]])/d - (((16*I)/35)*(e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/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 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 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 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 (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)} \, dx &=\left ((e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx\\ &=-\frac{2 i (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{\left (6 a (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac{1}{(e \sec (c+d x))^{3/2} \sqrt{a+i a \tan (c+d x)}} \, dx}{7 e^2}\\ &=\frac{12 i a (e \cos (c+d x))^{7/2} \sec ^2(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{\left (24 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{35 e^2}\\ &=\frac{12 i a (e \cos (c+d x))^{7/2} \sec ^2(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{16 i (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{\left (16 a (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}\right ) \int \frac{\sqrt{e \sec (c+d x)}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{35 e^4}\\ &=\frac{12 i a (e \cos (c+d x))^{7/2} \sec ^2(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}+\frac{32 i a (e \cos (c+d x))^{7/2} \sec ^4(c+d x)}{35 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i (e \cos (c+d x))^{7/2} \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{16 i (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}\\ \end{align*}

Mathematica [A]  time = 0.536439, size = 80, normalized size = 0.45 \[ \frac{e^3 \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)} (70 \sin (c+d x)+6 \sin (3 (c+d x))+35 i \cos (c+d x)+i \cos (3 (c+d x)))}{70 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.388, size = 97, normalized size = 0.5 \begin{align*}{\frac{2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+12\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +16\,i\cos \left ( dx+c \right ) +32\,\sin \left ( dx+c \right ) }{35\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35/d*(I*cos(d*x+c)^3+6*cos(d*x+c)^2*sin(d*x+c)+8*I*cos(d*x+c)+16*sin(d*x+c))*(e*cos(d*x+c))^(7/2)*(a*(I*sin(
d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3

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Maxima [A]  time = 2.92401, size = 273, normalized size = 1.53 \begin{align*} \frac{{\left (7 i \, e^{3} \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) - 5 i \, e^{3} \cos \left (\frac{7}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) - 35 i \, e^{3} \cos \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 105 i \, e^{3} \cos \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 7 \, e^{3} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, e^{3} \sin \left (\frac{7}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 35 \, e^{3} \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right ) + 105 \, e^{3} \sin \left (\frac{1}{5} \, \arctan \left (\sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ), \cos \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right )\right )\right )\right )} \sqrt{a} \sqrt{e}}{140 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.07285, size = 300, normalized size = 1.68 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{1}{2}}{\left (-5 i \, e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 35 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 105 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, e^{3}\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{5}{2} i \, d x - \frac{5}{2} i \, c\right )}}{140 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*sqrt(2)*sqrt(1/2)*(-5*I*e^3*e^(6*I*d*x + 6*I*c) - 35*I*e^3*e^(4*I*d*x + 4*I*c) + 105*I*e^3*e^(2*I*d*x +
2*I*c) + 7*I*e^3)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-5/2*I*d*x - 5/2*I*c)/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))**(7/2)*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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

[Out]

Timed out