3.4.46 \(\int \frac {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [346]
Optimal. Leaf size=193 \[ \frac {35 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} \sqrt {a} d}+\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 a d}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d} \]
[Out]
35/128*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d*2^(1/2)/a^(1/2)+35/96*I*cos(d*x+c)
/d/(a+I*a*tan(d*x+c))^(1/2)+1/4*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-35/64*I*cos(d*x+c)*(a+I*a*tan(d*x+c)
)^(1/2)/a/d-7/24*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/a/d
________________________________________________________________________________________
Rubi [A]
time = 0.19, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3583, 3578,
3571, 3570, 212} \begin {gather*} -\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 a d}+\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {35 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} \sqrt {a} d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(((35*I)/64)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*Sqrt[a]*d) + (((35
*I)/96)*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + ((I/4)*Cos[c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]]) -
(((35*I)/64)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (((7*I)/24)*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c +
d*x]])/(a*d)
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3570
Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(a/(b*f)), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]
Rule 3571
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/(2*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] && EqQ[m/2 + n, 0] && GtQ[n, 0]
Rule 3578
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 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 {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{8 a}\\ &=\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}+\frac {35}{48} \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}+\frac {35 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{64 a}\\ &=\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 a d}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}+\frac {35}{128} \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 a d}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}+\frac {(35 i) \text {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}\\ &=\frac {35 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{64 \sqrt {2} \sqrt {a} d}+\frac {35 i \cos (c+d x)}{96 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos ^3(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{64 a d}-\frac {7 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 a d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.75, size = 117, normalized size = 0.61 \begin {gather*} \frac {\sec (c+d x) \left (-41 i+105 i \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-43 i \cos (2 (c+d x))-2 i \cos (4 (c+d x))+133 \sin (2 (c+d x))+14 \sin (4 (c+d x))\right )}{384 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(Sec[c + d*x]*(-41*I + (105*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - (43*I)*C
os[2*(c + d*x)] - (2*I)*Cos[4*(c + d*x)] + 133*Sin[2*(c + d*x)] + 14*Sin[4*(c + d*x)]))/(384*d*Sqrt[a + I*a*Ta
n[c + d*x]])
________________________________________________________________________________________
Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 345 vs. \(2 (156 ) = 312\).
time = 0.94, size = 346, normalized size = 1.79
| | |
| method |
result |
size |
| | |
| default |
\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (192 i \left (\cos ^{5}\left (d x +c \right )\right )+192 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+105 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+105 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )+105 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+56 i \left (\cos ^{3}\left (d x +c \right )\right )+280 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-420 i \cos \left (d x +c \right )\right )}{768 d a}\) |
\(346\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/768/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(192*I*cos(d*x+c)^5+192*sin(d*x+c)*cos(d*x+c)^4+105*I*(
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)+105*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*
x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)+105*I*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(
1/2))*2^(1/2)+56*I*cos(d*x+c)^3+280*cos(d*x+c)^2*sin(d*x+c)-420*I*cos(d*x+c))/a
________________________________________________________________________________________
Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 1938 vs. \(2 (146) = 292\).
time = 0.69, size = 1938, normalized size = 10.04 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
-1/1536*(4*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d
*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(3/4)*((-3*I*sqrt(2)*cos(4*d*x + 4
*c) - 3*sqrt(2)*sin(4*d*x + 4*c) + 8*I*sqrt(2))*cos(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + (3*sqrt(2)*cos(4*d*x + 4*c) - 3*I*sqrt(2)
*sin(4*d*x + 4*c) - 8*sqrt(2))*sin(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*a
rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + 12*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4
*d*x + 4*c))) + 1)^(1/4)*((-I*sqrt(2)*cos(4*d*x + 4*c) - 12*I*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*
d*x + 4*c))) - sqrt(2)*sin(4*d*x + 4*c) - 12*sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 24
*I*sqrt(2))*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x +
4*c), cos(4*d*x + 4*c))) + 1)) + (sqrt(2)*cos(4*d*x + 4*c) + 12*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(
4*d*x + 4*c))) - I*sqrt(2)*sin(4*d*x + 4*c) - 12*I*sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))
) - 24*sqrt(2))*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*
x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + 105*(2*sqrt(2)*arctan2((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*
d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c),
cos(4*d*x + 4*c))) + 1)^(1/4)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*ar
ctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)), (cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + si
n(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))
+ 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4
*c), cos(4*d*x + 4*c))) + 1)) + 1) - 2*sqrt(2)*arctan2((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2
+ sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*
c))) + 1)^(1/4)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*
x + 4*c), cos(4*d*x + 4*c))) + 1)), (cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(
sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos
(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x
+ 4*c))) + 1)) - 1) - I*sqrt(2)*log(sqrt(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arc
tan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*cos(
1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x
+ 4*c))) + 1))^2 + sqrt(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4
*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*sin(1/2*arctan2(sin(1/
2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 +
2*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c
)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(
4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + 1) + I*sqrt(2)*
log(sqrt(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x
+ 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4
*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 + sqrt(cos(1/2*a
rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos
(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4
*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 - 2*(cos(1/2*arctan2(sin(4*d*x + 4
*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d
*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))),
cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + 1))*sqrt(a))/(a*d)
________________________________________________________________________________________
Fricas [A]
time = 0.38, size = 267, normalized size = 1.38 \begin {gather*} \frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\right ) + 105 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{32 \, d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 88 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 41 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 45 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{384 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/384*(-105*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(4*I*d*x + 4*I*c)*log(-35/32*(sqrt(2)*sqrt(1/2)*(I*d*e^(2*I*d*x
+ 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d) + 105*I*sqrt(1/2)*a
*d*sqrt(1/(a*d^2))*e^(4*I*d*x + 4*I*c)*log(-35/32*(sqrt(2)*sqrt(1/2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(
e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
*(-8*I*e^(8*I*d*x + 8*I*c) - 88*I*e^(6*I*d*x + 6*I*c) - 41*I*e^(4*I*d*x + 4*I*c) + 45*I*e^(2*I*d*x + 2*I*c) +
6*I))*e^(-4*I*d*x - 4*I*c)/(a*d)
________________________________________________________________________________________
Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3/(a+I*a*tan(d*x+c))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
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(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)^3/sqrt(I*a*tan(d*x + c) + a), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(1/2),x)
[Out]
int(cos(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(1/2), x)
________________________________________________________________________________________