3.4.60 \(\int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [360]
Optimal. Leaf size=157 \[ \frac {15 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d} \]
[Out]
15/64*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+5/16*I*cos(d*x+c)/a
/d/(a+I*a*tan(d*x+c))^(1/2)-15/32*I*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d+1/4*I*cos(d*x+c)/d/(a+I*a*tan(d*
x+c))^(3/2)
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Rubi [A]
time = 0.15, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3583, 3571,
3570, 212} \begin {gather*} \frac {15 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
(((15*I)/32)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) + ((I/4
)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((5*I)/16)*Cos[c + d*x])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])
- (((15*I)/32)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*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 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 (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{8 a}\\ &=\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}+\frac {15 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{32 a^2}\\ &=\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d}+\frac {15 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{64 a}\\ &=\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d}+\frac {(15 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 )}{32 a d}\\ &=\frac {15 i \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d}\\ \end {align*}
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Mathematica [A]
time = 1.23, size = 120, normalized size = 0.76 \begin {gather*} \frac {\sec (c+d x) \left (\frac {30 e^{2 i (c+d x)} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-2 (-9+6 \cos (2 (c+d x))+10 i \sin (2 (c+d x)))\right )}{64 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
(Sec[c + d*x]*((30*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])/Sqrt[1 + E^((2*I)*(c + d*x))] -
2*(-9 + 6*Cos[2*(c + d*x)] + (10*I)*Sin[2*(c + d*x)])))/(64*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]
])
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 345 vs. \(2 (126 ) = 252\).
time = 0.79, size = 346, normalized size = 2.20
| | |
| method |
result |
size |
| | |
| default |
\(\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (64 i \left (\cos ^{5}\left (d x +c \right )\right )+15 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )+64 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+8 i \left (\cos ^{3}\left (d x +c \right )\right )+15 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )+40 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-60 i \cos \left (d x +c \right )\right )}{128 d \,a^{2}}\) |
\(346\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/128/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(64*I*cos(d*x+c)^5+15*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)+64*sin(d*x+c)*cos(d*x+c)^4+15*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)+8*I*cos(d*x+c)^3+15*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)+40*cos(d*x+c)^2*sin(d*x+c)-60*I*cos(d*x+c))/a^2
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 1821 vs. \(2 (118) = 236\).
time = 0.65, size = 1821, normalized size = 11.60 \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)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
-1/256*(36*(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)*((-I*sqrt(2)*cos(4*d*x + 4*c
) - sqrt(2)*sin(4*d*x + 4*c))*cos(3/2*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)) + (sqrt(2)*cos(4*d*x + 4*c) - I*sqrt(2)*sin(4*d*x + 4*c))*sin
(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)))*sqrt(a) + 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)^(1/4)*((7*I*sqrt
(2)*cos(4*d*x + 4*c) + 7*sqrt(2)*sin(4*d*x + 4*c) + 8*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)) - (7*sqrt(2)*cos(4*d*x + 4
*c) - 7*I*sqrt(2)*sin(4*d*x + 4*c) + 8*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) + 15*(2*sqrt(2)*arctan2((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)^(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) - 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(s
in(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), co
s(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*arct
an2(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*arcta
n2(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(si
n(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*arcta
n2(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))*sqrt(a))/(a^2
*d)
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 270 vs. \(2 (118) = 236\).
time = 0.37, size = 270, normalized size = 1.72 \begin {gather*} \frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, a d}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, a d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{64 \, a^{2} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
1/64*(-15*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-15/16*(sqrt(2)*sqrt(1/2)*(I*a*d*e^(2*I*
d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) + 15*I*
sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-15/16*(sqrt(2)*sqrt(1/2)*(-I*a*d*e^(2*I*d*x + 2*I*c
) - I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) + sqrt(2)*sqrt(a/(
e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c) + 11*I*e^(2*I*d*x + 2*I*c) + 2*I))
*e^(-4*I*d*x - 4*I*c)/(a^2*d)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\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(cos(d*x+c)/(a+I*a*tan(d*x+c))**(3/2),x)
[Out]
Integral(cos(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(cos(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)/(I*a*tan(d*x + c) + a)^(3/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\cos \left (c+d\,x\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)/(a + a*tan(c + d*x)*1i)^(3/2),x)
[Out]
int(cos(c + d*x)/(a + a*tan(c + d*x)*1i)^(3/2), x)
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