3.4.39 \(\int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [339]

Optimal. Leaf size=292 \[ -\frac {429 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} \sqrt {a} d}+\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-429/2048*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)+429/1024*I/d/(a+I*a*tan(d*
x+c))^(1/2)+429/896*I*a^3/d/(a+I*a*tan(d*x+c))^(7/2)-1/6*I*a^6/d/(a-I*a*tan(d*x+c))^3/(a+I*a*tan(d*x+c))^(7/2)
-13/48*I*a^5/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(7/2)-143/192*I*a^4/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x
+c))^(7/2)+429/1280*I*a^2/d/(a+I*a*tan(d*x+c))^(5/2)+143/512*I*a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.12, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \begin {gather*} -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}}-\frac {429 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} \sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-429*I)/1024)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + (((429*I)/896)*a
^3)/(d*(a + I*a*Tan[c + d*x])^(7/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(7/2)) -
(((13*I)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(7/2)) - (((143*I)/192)*a^4)/(d*(a - I*a
*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(7/2)) + (((429*I)/1280)*a^2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((143*
I)/512)*a)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((429*I)/1024)/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\cos ^6(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {\left (i a^7\right ) \text {Subst}\left (\int \frac {1}{(a-x)^4 (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {\left (13 i a^6\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{12 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {\left (143 i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{96 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}-\frac {\left (429 i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}-\frac {\left (429 i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}-\frac {\left (429 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{512 d}\\ &=\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}-\frac {(429 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{1024 d}\\ &=\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}}-\frac {(429 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{2048 d}\\ &=\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}}-\frac {(429 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{1024 d}\\ &=-\frac {429 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{1024 \sqrt {2} \sqrt {a} d}+\frac {429 i a^3}{896 d (a+i a \tan (c+d x))^{7/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{7/2}}-\frac {13 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{7/2}}-\frac {143 i a^4}{192 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{7/2}}+\frac {429 i a^2}{1280 d (a+i a \tan (c+d x))^{5/2}}+\frac {143 i a}{512 d (a+i a \tan (c+d x))^{3/2}}+\frac {429 i}{1024 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.20, size = 178, normalized size = 0.61 \begin {gather*} -\frac {i e^{-6 i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-240-2064 e^{2 i (c+d x)}-9008 e^{4 i (c+d x)}-40784 e^{6 i (c+d x)}+13755 e^{8 i (c+d x)}+2590 e^{10 i (c+d x)}+280 e^{12 i (c+d x)}\right )+45045 e^{7 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{107520 d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-1/107520*I)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-240 - 2064*E^((2*I)*(c + d*x)) - 9008*E^((4*I)*(c + d*x)) - 40
784*E^((6*I)*(c + d*x)) + 13755*E^((8*I)*(c + d*x)) + 2590*E^((10*I)*(c + d*x)) + 280*E^((12*I)*(c + d*x))) +
45045*E^((7*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))]))/(d*E^((6*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt
[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 1.05, size = 389, normalized size = 1.33

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-61440 i \left (\cos ^{8}\left (d x +c \right )\right )-61440 \sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )-6656 i \left (\cos ^{6}\left (d x +c \right )\right )-73216 \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )+45045 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-13728 i \left (\cos ^{4}\left (d x +c \right )\right )+45045 \sin \left (d x +c \right ) \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+45045 i \arctan \left (\frac {\left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )+i\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-96096 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-60060 i \left (\cos ^{2}\left (d x +c \right )\right )-180180 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{430080 d a}\)	\(389\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/430080/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-61440*I*cos(d*x+c)^8-61440*sin(d*x+c)*cos(d*x+c)^
7-6656*I*cos(d*x+c)^6-73216*sin(d*x+c)*cos(d*x+c)^5+45045*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d
*x+c)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-13728*I
*cos(d*x+c)^4+45045*sin(d*x+c)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)+45045*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^
(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-96096*s
in(d*x+c)*cos(d*x+c)^3-60060*I*cos(d*x+c)^2-180180*sin(d*x+c)*cos(d*x+c))/a

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Maxima [A]
time = 0.49, size = 246, normalized size = 0.84 \begin {gather*} \frac {i \, {\left (45045 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (45045 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a - 240240 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 396396 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 164736 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 36608 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 19968 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 15360 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3}}\right )}}{430080 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/430080*I*(45045*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(
I*a*tan(d*x + c) + a))) + 4*(45045*(I*a*tan(d*x + c) + a)^6*a - 240240*(I*a*tan(d*x + c) + a)^5*a^2 + 396396*(
I*a*tan(d*x + c) + a)^4*a^3 - 164736*(I*a*tan(d*x + c) + a)^3*a^4 - 36608*(I*a*tan(d*x + c) + a)^2*a^5 - 19968
*(I*a*tan(d*x + c) + a)*a^6 - 15360*a^7)/((I*a*tan(d*x + c) + a)^(13/2) - 6*(I*a*tan(d*x + c) + a)^(11/2)*a +
12*(I*a*tan(d*x + c) + a)^(9/2)*a^2 - 8*(I*a*tan(d*x + c) + a)^(7/2)*a^3))/(a*d)

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Fricas [A]
time = 0.39, size = 315, normalized size = 1.08 \begin {gather*} \frac {{\left (-45045 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 45045 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a 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 (-280 i \, e^{\left (14 i \, d x + 14 i \, c\right )} - 2870 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 16345 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 27029 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 49792 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11072 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2304 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 240 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{215040 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/215040*(-45045*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(7*I*d*x + 7*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a*d*e^(2*I*d*x
+ 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 450
45*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(7*I*d*x + 7*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*
d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - a*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))*(-280*I*e^(14*I*d*x + 14*I*c) - 2870*I*e^(12*I*d*x + 12*I*c) - 16345*I*e^(10*I*d*x +
10*I*c) + 27029*I*e^(8*I*d*x + 8*I*c) + 49792*I*e^(6*I*d*x + 6*I*c) + 11072*I*e^(4*I*d*x + 4*I*c) + 2304*I*e^
(2*I*d*x + 2*I*c) + 240*I))*e^(-7*I*d*x - 7*I*c)/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{6}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**6/sqrt(I*a*(tan(c + d*x) - I)), 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)^6/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^6/sqrt(I*a*tan(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^6}{\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)^6/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cos(c + d*x)^6/(a + a*tan(c + d*x)*1i)^(1/2), x)

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