3.338 \(\int \frac {\cos ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=219 \[ -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d} \]

[Out]

-63/256*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)+63/128*I/d/(a+I*a*tan(d*x+c)
)^(1/2)+63/160*I*a^2/d/(a+I*a*tan(d*x+c))^(5/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(5/2)-9/16
*I*a^3/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2)+21/64*I*a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.12, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-63*I)/128)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + (((63*I)/160)*a^2)
/(d*(a + I*a*Tan[c + d*x])^(5/2)) - ((I/4)*a^4)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(5/2)) - ((
(9*I)/16)*a^3)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (((21*I)/64)*a)/(d*(a + I*a*Tan[c + d
*x])^(3/2)) + ((63*I)/128)/(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 ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {\left (i a^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (9 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (63 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (63 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}-\frac {(63 i a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {(63 i) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}-\frac {(63 i) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{128 d}\\ &=-\frac {63 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} \sqrt {a} d}+\frac {63 i a^2}{160 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {9 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {21 i a}{64 d (a+i a \tan (c+d x))^{3/2}}+\frac {63 i}{128 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 152, normalized size = 0.69 \[ -\frac {i e^{-4 i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-56 e^{2 i (c+d x)}-288 e^{4 i (c+d x)}+85 e^{6 i (c+d x)}+10 e^{8 i (c+d x)}-8\right )+315 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{640 d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-1/640*I)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-8 - 56*E^((2*I)*(c + d*x)) - 288*E^((4*I)*(c + d*x)) + 85*E^((6*I
)*(c + d*x)) + 10*E^((8*I)*(c + d*x))) + 315*E^((5*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))]))/(d*E^((4*I)*(c + d
*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 0.71, size = 293, normalized size = 1.34 \[ \frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 315 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-10 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 95 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 203 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 344 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 64 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{1280 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/1280*(-315*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(5*I*d*x + 5*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)) + 315*I*s
qrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqr
t(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))*(-10*I*e^(10*I*d*x + 10*I*c) - 95*I*e^(8*I*d*x + 8*I*c) + 203*I*e^(6*I*d*x + 6*I*c) + 344*
I*e^(4*I*d*x + 4*I*c) + 64*I*e^(2*I*d*x + 2*I*c) + 8*I))*e^(-5*I*d*x - 5*I*c)/(a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^4/sqrt(I*a*tan(d*x + c) + a), x)

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maple [B]  time = 1.15, size = 368, normalized size = 1.68 \[ \frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (512 i \left (\cos ^{6}\left (d x +c \right )\right )+512 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+315 i \cos \left (d x +c \right ) \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+96 i \left (\cos ^{4}\left (d x +c \right )\right )+315 i \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+315 \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+672 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+420 i \left (\cos ^{2}\left (d x +c \right )\right )+1260 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{2560 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

1/2560/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(512*I*cos(d*x+c)^6+512*cos(d*x+c)^5*sin(d*x+c)+315*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)*cos(d*x+c)+96*I*cos(d*x+c)^4+315*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)+315*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)+672*cos(d*x+c)^3*sin(d*x+c)+420*I*cos(d*x+c)^2+1260*cos(d*x+c)*sin(d*x+c))/a

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maxima [A]  time = 0.84, size = 192, normalized size = 0.88 \[ \frac {i \, {\left (315 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1050 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} + 672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 192 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 128 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}\right )}}{2560 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/2560*I*(315*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*(315*(I*a*tan(d*x + c) + a)^4*a - 1050*(I*a*tan(d*x + c) + a)^3*a^2 + 672*(I*a*tan(d*x
+ c) + a)^2*a^3 + 192*(I*a*tan(d*x + c) + a)*a^4 + 128*a^5)/((I*a*tan(d*x + c) + a)^(9/2) - 4*(I*a*tan(d*x +
c) + a)^(7/2)*a + 4*(I*a*tan(d*x + c) + a)^(5/2)*a^2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cos(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**4/sqrt(I*a*(tan(c + d*x) - I)), x)

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