3.351 \(\int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=175 \[ -\frac {7 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 i}{16 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-7/32*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+7/16*I/a/d/(a+I*a*tan(d*x+c))^
(1/2)+7/20*I*a/d/(a+I*a*tan(d*x+c))^(5/2)-1/2*I*a^2/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2)+7/24*I/d/(a+
I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.12, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {7 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}+\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 i}{16 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-7*I)/16)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) + (((7*I)/20)*a)/(d*(a
+ I*a*Tan[c + d*x])^(5/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + ((7*I)/24)
/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ((7*I)/16)/(a*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 51

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 63

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 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])

Rule 3487

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 ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (7 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {(7 i a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}-\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{16 d}\\ &=\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 i}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{32 a d}\\ &=\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 i}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(7 i) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{16 a d}\\ &=-\frac {7 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}+\frac {7 i a}{20 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {7 i}{24 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 i}{16 a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.00, size = 142, normalized size = 0.81 \[ -\frac {i e^{-5 i (c+d x)} \sec (c+d x) \left (-38 e^{2 i (c+d x)}-148 e^{4 i (c+d x)}-101 e^{6 i (c+d x)}+15 e^{8 i (c+d x)}+105 e^{5 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-6\right )}{480 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-1/480*I)*(-6 - 38*E^((2*I)*(c + d*x)) - 148*E^((4*I)*(c + d*x)) - 101*E^((6*I)*(c + d*x)) + 15*E^((8*I)*(c
+ d*x)) + 105*E^((5*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x])/(a*d*E
^((5*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.47, size = 294, normalized size = 1.68 \[ \frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 105 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-15 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 101 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 38 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{480 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/480*(-105*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x
+ 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
+ 105*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*
I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqr
t(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-15*I*e^(8*I*d*x + 8*I*c) + 101*I*e^(6*I*d*x + 6*I*c) + 148*I*e^(4*I*d
*x + 4*I*c) + 38*I*e^(2*I*d*x + 2*I*c) + 6*I))*e^(-5*I*d*x - 5*I*c)/(a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(I*a*tan(d*x + c) + a)^(3/2), x)

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maple [B]  time = 1.07, size = 368, normalized size = 2.10 \[ \frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (384 i \left (\cos ^{6}\left (d x +c \right )\right )+384 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+32 i \left (\cos ^{4}\left (d x +c \right )\right )+105 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+105 i \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+224 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+105 \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+140 i \left (\cos ^{2}\left (d x +c \right )\right )+420 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{960 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/960/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(384*I*cos(d*x+c)^6+384*cos(d*x+c)^5*sin(d*x+c)+32*I*co
s(d*x+c)^4+105*I*cos(d*x+c)*(-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)+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)+224*cos(d*x+c)
^3*sin(d*x+c)+105*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)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+140*I*cos(d*x+c)^2+420*cos(d*x+c)*sin(d*x+c))/a^2

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maxima [A]  time = 0.50, size = 153, normalized size = 0.87 \[ \frac {i \, {\left (\frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 140 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 56 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 48 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}\right )}}{960 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/960*I*(105*sqrt(2)*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)))/sqrt(a) + 4*(105*(I*a*tan(d*x + c) + a)^3 - 140*(I*a*tan(d*x + c) + a)^2*a - 56*(I*a*tan(d*x + c) +
a)*a^2 - 48*a^3)/((I*a*tan(d*x + c) + a)^(7/2) - 2*(I*a*tan(d*x + c) + a)^(5/2)*a))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(cos(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(3/2), x)

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