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

Optimal. Leaf size=193 \[ -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a}{64 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d} \]

[Out]

-35/128*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^(1/2)/d*2^(1/2)+35/64*I*a/d/(a+I*a*tan(d*x+c
))^(1/2)+35/96*I*a^2/d/(a+I*a*tan(d*x+c))^(3/2)-1/4*I*a^4/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(3/2)-7/16
*I*a^3/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.11, antiderivative size = 193, 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} \[ -\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a}{64 d \sqrt {a+i a \tan (c+d x)}}-\frac {35 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-35*I)/64)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/96)*a^2)/(
d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I/4)*a^4)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(3/2)) - (((7
*I)/16)*a^3)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + (((35*I)/64)*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 \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)^{5/2}} \, 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))^{3/2}}-\frac {\left (7 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{5/2}} \, 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))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (35 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (35 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a}{64 d \sqrt {a+i a \tan (c+d x)}}-\frac {(35 i a) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a}{64 d \sqrt {a+i a \tan (c+d x)}}-\frac {(35 i a) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{64 d}\\ &=-\frac {35 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {35 i a^2}{96 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^4}{4 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac {7 i a^3}{16 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {35 i a}{64 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 133, normalized size = 0.69 \[ -\frac {i e^{-4 i (c+d x)} \left (-88 e^{2 i (c+d x)}-41 e^{4 i (c+d x)}+45 e^{6 i (c+d x)}+6 e^{8 i (c+d x)}+105 e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-8\right ) \sqrt {a+i a \tan (c+d x)}}{384 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/384*I)*(-8 - 88*E^((2*I)*(c + d*x)) - 41*E^((4*I)*(c + d*x)) + 45*E^((6*I)*(c + d*x)) + 6*E^((8*I)*(c + d
*x)) + 105*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*
x]])/(d*E^((4*I)*(c + d*x)))

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fricas [A]  time = 1.39, size = 274, normalized size = 1.42 \[ \frac {{\left (105 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {1}{32} \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (128 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 128 \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 105 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {1}{32} \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-128 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 128 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 128 \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-6 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 45 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 41 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{384 \, d} \]

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]

1/384*(105*sqrt(1/2)*d*sqrt(-a/d^2)*e^(3*I*d*x + 3*I*c)*log(1/32*(sqrt(2)*sqrt(1/2)*(128*I*d*e^(2*I*d*x + 2*I*
c) + 128*I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + 128*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 105*
sqrt(1/2)*d*sqrt(-a/d^2)*e^(3*I*d*x + 3*I*c)*log(1/32*(sqrt(2)*sqrt(1/2)*(-128*I*d*e^(2*I*d*x + 2*I*c) - 128*I
*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + 128*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))*(-6*I*e^(8*I*d*x + 8*I*c) - 45*I*e^(6*I*d*x + 6*I*c) + 41*I*e^(4*I*d*x + 4*I*c) +
88*I*e^(2*I*d*x + 2*I*c) + 8*I))*e^(-3*I*d*x - 3*I*c)/d

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

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

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maple [B]  time = 1.35, size = 741, normalized size = 3.84 \[ \text {result too large to display} \]

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]

-1/3072/d*(128*I*cos(d*x+c)^7-105*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)*arctan(1/2*(-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)))^(7/2)-1680*I*cos(d*x+c)^4-315*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*arct
an(1/2*(-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)))^(7/2)+224*I*cos(d*x+c)^6-3
15*cos(d*x+c)*2^(1/2)*sin(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(7/2)+560*I*cos(d*x+c)^5-105*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)-315*I*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c))
)^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))-105*I*cos(d*x+c)^3*2^(
1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)/cos(d*x+c)*2^(1/2))-768*sin(d*x+c)*cos(d*x+c)^7-105*I*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+896*sin(d*x+c)*cos(d*x+c)^6
-315*I*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))-1120*cos(d*x+c)^5*sin(d*x+c)+768*I*cos(d*x+c)^8+1680*sin(d*x+c)*co
s(d*x+c)^4)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3

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maxima [A]  time = 0.85, size = 176, normalized size = 0.91 \[ \frac {i \, {\left (105 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 350 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 224 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 64 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}}\right )}}{768 \, a d} \]

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]

1/768*I*(105*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*t
an(d*x + c) + a))) + 4*(105*(I*a*tan(d*x + c) + a)^3*a^2 - 350*(I*a*tan(d*x + c) + a)^2*a^3 + 224*(I*a*tan(d*x
+ c) + a)*a^4 + 64*a^5)/((I*a*tan(d*x + c) + a)^(7/2) - 4*(I*a*tan(d*x + c) + a)^(5/2)*a + 4*(I*a*tan(d*x + c
) + a)^(3/2)*a^2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

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 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cos ^{4}{\left (c + d x \right )}\, dx \]

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

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