3.311 \(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=137 \[ -\frac {3 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{4 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{16 d (a-i a \tan (c+d x))} \]

[Out]

-3/32*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)-1/4*I*a^4*(a+I*a*tan(d*x+c))^(
1/2)/d/(a-I*a*tan(d*x+c))^2-3/16*I*a^3*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{4 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{16 d (a-i a \tan (c+d x))}-\frac {3 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.84, size = 116, normalized size = 0.85 \[ -\frac {i a^2 e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (5+2 e^{2 i (c+d x)}\right )+3 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-1/32*I)*a^2*Sqrt[1 + E^((2*I)*(c + d*x))]*(E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*(5 + 2*E^((2*I)*(c
+ d*x))) + 3*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x)))

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fricas [B]  time = 0.47, size = 262, normalized size = 1.91 \[ \frac {3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {{\left (32 \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (32 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2}}\right ) - 3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {{\left (32 \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-32 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 32 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2}}\right ) + \sqrt {2} {\left (-2 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/32*(3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(1/8*(32*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(32*I*d*
e^(2*I*d*x + 2*I*c) + 32*I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 3*sqrt(1/2)*sqrt(-a^5
/d^2)*d*log(1/8*(32*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(-32*I*d*e^(2*I*d*x + 2*I*c) - 32*I
*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + sqrt(2)*(-2*I*a^2*e^(5*I*d*x + 5*I*c) - 7*I*a^2
*e^(3*I*d*x + 3*I*c) - 5*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.34, size = 744, normalized size = 5.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/256/d*(128*I*cos(d*x+c)^7+48*I*cos(d*x+c)^4+3*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+9*I*sin(d*x+c)*cos(d*x+c)^2*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)+9*c
os(d*x+c)^2*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)+3*I*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^(1/2)*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+9*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+cos(d*x+c)))^(7/2)+3*2^(1/2)*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)*sin(d*x+c)+96*I*cos(d*x+c)^6-16*I*cos(d*x+c)^5+256*sin(
d*x+c)*cos(d*x+c)^7+3*I*sin(d*x+c)*cos(d*x+c)^3*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)-128*sin(d*x+c)*cos(d*x+c)^6-256*I*cos(d*x+c)^8+
32*cos(d*x+c)^5*sin(d*x+c)+9*I*sin(d*x+c)*cos(d*x+c)*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)-48*sin(d*x+c)*cos(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*a^2

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maxima [A]  time = 0.62, size = 140, normalized size = 1.02 \[ \frac {i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 10 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}\right )}}{64 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/64*I*(3*sqrt(2)*a^(7/2)*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))) + 4*(3*(I*a*tan(d*x + c) + a)^(3/2)*a^4 - 10*sqrt(I*a*tan(d*x + c) + a)*a^5)/((I*a*tan(d*x + c
) + a)^2 - 4*(I*a*tan(d*x + c) + a)*a + 4*a^2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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