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*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]])

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Rubi [A]  time = 0.112087, antiderivative size = 193, normalized size of antiderivative = 1., 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 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) \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*}

Mathematica [A]  time = 0.370383, 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]

((-I/384)*(-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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Maple [B]  time = 0.355, size = 741, normalized size = 3.8 \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))^(1/2),x)

[Out]

1/3072/d*(-768*I*cos(d*x+c)^8-224*I*cos(d*x+c)^6+105*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)+105*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))*cos(d*x+c)^3*sin(d*x
+c)+315*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)+105*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)+315*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)+105*2^(1/2)*arctan(1/2*2^(1/2)*(-2*c
os(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)+315*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))*cos(d*x+
c)*sin(d*x+c)-128*I*cos(d*x+c)^7+768*sin(d*x+c)*cos(d*x+c)^7-560*I*cos(d*x+c)^5-896*cos(d*x+c)^6*sin(d*x+c)+16
80*I*cos(d*x+c)^4+1120*cos(d*x+c)^5*sin(d*x+c)+315*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))*cos(d*x+c)^2*sin(d*x+c)-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 [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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.84487, size = 868, normalized size = 4.5 \begin{align*} \frac{{\left (105 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{1}{35} \,{\left (70 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 35 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 (4 i \, d x + 4 i \, c\right )} \log \left (\frac{1}{35} \,{\left (-70 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 35 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{384 \, 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))^(1/2),x, algorithm="fricas")

[Out]

1/384*(105*sqrt(1/2)*d*sqrt(-a/d^2)*e^(4*I*d*x + 4*I*c)*log(1/35*(70*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2
*I*c) + 35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*
c)) - 105*sqrt(1/2)*d*sqrt(-a/d^2)*e^(4*I*d*x + 4*I*c)*log(1/35*(-70*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2
*I*c) + 35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*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^(I*d*x + I*c))*e^(-4*I*d*x - 4*I*c)/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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4}\,{d x} \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))^(1/2),x, algorithm="giac")

[Out]

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