3.346 \(\int \frac{\cos ^3(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)
Optimal. Leaf size=193 \[ -\frac{7 i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 a d}+\frac{i \cos ^3(c+d x)}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{35 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 a d}+\frac{35 i \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}+\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{64 \sqrt{2} \sqrt{a} d} \]
[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)
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Rubi [A] time = 0.266155, antiderivative size = 193, normalized size of antiderivative = 1.,
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 =
{3502, 3497, 3490, 3489, 206} \[ -\frac{7 i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{24 a d}+\frac{i \cos ^3(c+d x)}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{35 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{64 a d}+\frac{35 i \cos (c+d x)}{96 d \sqrt{a+i a \tan (c+d x)}}+\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{64 \sqrt{2} \sqrt{a} d} \]
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 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 3490
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 3489
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
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) \operatorname{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 = 0.527769, size = 117, normalized size = 0.61 \[ \frac{\sec (c+d x) \left (133 \sin (2 (c+d x))+14 \sin (4 (c+d x))-43 i \cos (2 (c+d x))-2 i \cos (4 (c+d x))+105 i \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-41 i\right )}{384 d \sqrt{a+i a \tan (c+d x)}} \]
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]])
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Maple [B] time = 0.329, size = 346, normalized size = 1.8 \begin{align*}{\frac{1}{768\,ad}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 192\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}+105\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \cos \left ( dx+c \right ) +192\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+105\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) +56\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+105\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) +280\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -420\,i\cos \left ( dx+c \right ) \right ) } \end{align*}
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)
[Out]
1/768/d/a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(192*I*cos(d*x+c)^5+105*I*2^(1/2)*(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2))*cos(d*x+c)+192*sin(d*x+c)*cos(d*x+c)^4+105*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/
2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+56*I*cos(d*x+c)^3+105*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)+280*cos(d*x+c)^2*sin(d*x+c)-420*I*cos(d*x+c))
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Maxima [B] time = 2.63402, size = 2618, normalized size = 13.56 \begin{align*} \text{result too large to display} \end{align*}
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*((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)*((12*I*sqrt(2)*cos(4*d*x + 4*c)
+ 12*sqrt(2)*sin(4*d*x + 4*c) - 32*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)) - 4*(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*
arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + (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)*((12*I*sqrt(2)*cos(4*d*x + 4*c) + 144*I*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4
*d*x + 4*c))) + 12*sqrt(2)*sin(4*d*x + 4*c) + 144*sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))
- 288*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)) - (12*sqrt(2)*cos(4*d*x + 4*c) + 144*sqrt(2)*cos(1/2*arctan2(sin(4*d*x +
4*c), cos(4*d*x + 4*c))) - 12*I*sqrt(2)*sin(4*d*x + 4*c) - 144*I*sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos
(4*d*x + 4*c))) - 288*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*a
rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) - (210*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) - 210*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*a
rctan2(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 + s
in(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) - 105*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*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*ar
ctan2(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(s
in(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) + 105*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*arctan
2(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*d)
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Fricas [B] time = 2.03375, size = 865, normalized size = 4.48 \begin{align*} \frac{{\left (105 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 105 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\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 )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{384 \, a d} \end{align*}
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^(5*I*d*x + 5*I*c)*log((2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(I*d*x +
I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
- 105*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(5*I*d*x + 5*I*c)*log(-(2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(I*d*x + I*
c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) +
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^(I*d*x + I*c))*e^(-5*I*d*x - 5*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(cos(d*x+c)**3/(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{\cos \left (d x + c\right )^{3}}{\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(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)