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

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

[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]))

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Rubi [A]  time = 0.0949402, antiderivative size = 137, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.704104, 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]

((-I/32)*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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Maple [B]  time = 0.336, size = 744, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.4793, size = 761, normalized size = 5.55 \begin{align*} \frac{\sqrt{2}{\left (-2 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 3 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d \log \left (\frac{{\left (6 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{2}}\right ) - 3 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d \log \left (\frac{{\left (-6 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{2}}\right )}{32 \, d} \end{align*}

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*(sqrt(2)*(-2*I*a^2*e^(4*I*d*x + 4*I*c) - 7*I*a^2*e^(2*I*d*x + 2*I*c) - 5*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*e^(I*d*x + I*c) + 3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(1/3*(6*I*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2
*I*c) + 3*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d
*x - 2*I*c)/a^2) - 3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(1/3*(-6*I*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c) +
3*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*
I*c)/a^2))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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