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*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]])

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Rubi [A]  time = 0.12402, antiderivative size = 219, normalized size of antiderivative = 1., 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 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 \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*}

Mathematica [A]  time = 0.627008, 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]

((-I/640)*(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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Maple [B]  time = 0.361, size = 368, normalized size = 1.7 \begin{align*}{\frac{1}{2560\,ad}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+512\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +315\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \cos \left ( dx+c \right ) +96\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+315\,i\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) +315\,\sqrt{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +672\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +420\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1260\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \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]

1/2560/d/a*(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^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+96*I*cos(d*x+c)^4+315*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+315*2^(1/2)
*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+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))

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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 [A]  time = 2.57692, size = 922, normalized size = 4.21 \begin{align*} \frac{{\left (-315 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 ) + 315 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \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 (-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 )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{1280 \, a 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]

1/1280*(-315*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d
*x + 2*I*c) + 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)) + 315*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(6*I*d*x + 6*I*c)*log(-(2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(2*I*
d*x + 2*I*c) - 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))*(-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^(I*d*x + I*c))*e^(-6
*I*d*x - 6*I*c)/(a*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 \frac{\cos \left (d x + c\right )^{4}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{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(cos(d*x + c)^4/sqrt(I*a*tan(d*x + c) + a), x)