3.383 \(\int \frac{\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=233 \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 i}{64 a^3 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(((-11*I)/64)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(7/2)*d) + (((11*I)/36)*a)/(d*
(a + I*a*Tan[c + d*x])^(9/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + ((11*I)/
56)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + ((11*I)/80)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*I)/96)/(a^2*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + ((11*I)/64)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.142192, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 i}{64 a^3 d \sqrt{a+i a \tan (c+d x)}}-\frac{11 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-11*I)/64)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(7/2)*d) + (((11*I)/36)*a)/(d*
(a + I*a*Tan[c + d*x])^(9/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + ((11*I)/
56)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + ((11*I)/80)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*I)/96)/(a^2*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + ((11*I)/64)/(a^3*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 ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac{\left (11 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac{(11 i a) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}-\frac{(11 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{16 d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}-\frac{(11 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{32 a d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{(11 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{64 a^2 d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 i}{64 a^3 d \sqrt{a+i a \tan (c+d x)}}-\frac{(11 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{128 a^3 d}\\ &=\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 i}{64 a^3 d \sqrt{a+i a \tan (c+d x)}}-\frac{(11 i) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{64 a^3 d}\\ &=-\frac{11 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac{11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac{11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac{11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 i}{64 a^3 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.51676, size = 176, normalized size = 0.76 \[ -\frac{i e^{-11 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \sec ^3(c+d x) \left (\sqrt{1+e^{2 i (c+d x)}} \left (-460 e^{2 i (c+d x)}-1338 e^{4 i (c+d x)}-2416 e^{6 i (c+d x)}-4618 e^{8 i (c+d x)}+315 e^{10 i (c+d x)}-70\right )+3465 e^{9 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{161280 a^3 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-I/161280)*(1 + E^((2*I)*(c + d*x)))^(5/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-70 - 460*E^((2*I)*(c + d*x)) - 1
338*E^((4*I)*(c + d*x)) - 2416*E^((6*I)*(c + d*x)) - 4618*E^((8*I)*(c + d*x)) + 315*E^((10*I)*(c + d*x))) + 34
65*E^((9*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^3)/(a^3*d*E^((11*I)*(c + d*x))*Sqrt[a + I*a*Tan[
c + d*x]])

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Maple [B]  time = 0.365, size = 422, normalized size = 1.8 \begin{align*}{\frac{1}{80640\,{a}^{4}d}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 71680\,i \left ( \cos \left ( dx+c \right ) \right ) ^{10}+71680\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{9}-43520\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}-7680\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+512\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+5632\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +3465\,i\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 ) \sqrt{2}+1056\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+3465\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}\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 ) +3465\,\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 ) +7392\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +4620\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+13860\,\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)^2/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

1/80640/d/a^4*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(71680*I*cos(d*x+c)^10+71680*sin(d*x+c)*cos(d*x+c
)^9-43520*I*cos(d*x+c)^8-7680*sin(d*x+c)*cos(d*x+c)^7+512*I*cos(d*x+c)^6+5632*cos(d*x+c)^5*sin(d*x+c)+3465*I*(
-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))*cos(d*x+c)*2^(1/2)+1056*I*cos(d*x+c)^4+3465*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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))+3465*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)+7392*cos(d*x+c)^3*sin(d*x+c)+4620*I*cos(d*x+c)^2+13860*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)^2/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.74869, size = 1010, normalized size = 4.33 \begin{align*} \frac{{\left (-3465 i \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} 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 ) + 3465 i \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{4} d \sqrt{\frac{1}{a^{7} 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 (-315 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 4303 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 7034 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3754 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1798 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 530 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 70 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{40320 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/40320*(-3465*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c)*log((2*sqrt(1/2)*a^4*d*sqrt(1/(a^7*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)) + 3465*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(10*I*d*x + 10*I*c)*log(-(2*sqrt(1/2)*a^4*d*sq
rt(1/(a^7*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))*(-315*I*e^(12*I*d*x + 12*I*c) + 430
3*I*e^(10*I*d*x + 10*I*c) + 7034*I*e^(8*I*d*x + 8*I*c) + 3754*I*e^(6*I*d*x + 6*I*c) + 1798*I*e^(4*I*d*x + 4*I*
c) + 530*I*e^(2*I*d*x + 2*I*c) + 70*I)*e^(I*d*x + I*c))*e^(-10*I*d*x - 10*I*c)/(a^4*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)**2/(a+I*a*tan(d*x+c))**(7/2),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 )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(I*a*tan(d*x + c) + a)^(7/2), x)