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

Optimal. Leaf size=116 \[ -\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}-\frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{3 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

[Out]

((3*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((3*I)*a^3*Sqrt[a + I*a*Tan[
c + d*x]])/d - (I*a^3*(a + I*a*Tan[c + d*x])^(3/2))/(d*(a - I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0884955, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3487, 47, 50, 63, 206} \[ -\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}-\frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{3 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((3*I)*a^3*Sqrt[a + I*a*Tan[
c + d*x]])/d - (I*a^3*(a + I*a*Tan[c + d*x])^(3/2))/(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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 ^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{(a+x)^{3/2}}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac{\left (3 i a^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=-\frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac{\left (3 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}+\frac{\left (6 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{3 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{i a^3 (a+i a \tan (c+d x))^{3/2}}{d (a-i a \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.24282, size = 137, normalized size = 1.18 \[ -\frac{i \sqrt{2} e^{-4 i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (3 e^{i (c+d x)}+e^{3 i (c+d x)}-3 \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) (a+i a \tan (c+d x))^{7/2}}{d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(3*E^(I*(c + d*x)) + E^((3*I)*(c + d*x)) - 3*Sqr
t[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*(a + I*a*Tan[c + d*x])^(7/2))/(d*E^((4*I)*(c + d*x))*Sec[
c + d*x]^(7/2))

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Maple [B]  time = 0.374, size = 412, normalized size = 3.6 \begin{align*} -{\frac{{a}^{3}}{2\,d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 3\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}+3\,i\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}\sin \left ( dx+c \right ) +3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \sqrt{2}+3\,\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) +8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -4\,i\cos \left ( dx+c \right ) -4\,\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]

-1/2/d*a^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*I*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(3/2)*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^(1/2)+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
))^(3/2)*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctan(1/2*2^(1/2)*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+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))^(3/2)*sin(d*x+c)+8*I*cos(d*x+c)^4-4*I*cos(d*x+c)^3-8*cos(d*x+c)^3*sin(d*x+c)+4*cos(d*x
+c)^2*sin(d*x+c)-4*I*cos(d*x+c)-4*cos(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(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 [B]  time = 2.15239, size = 709, normalized size = 6.11 \begin{align*} \frac{\sqrt{2}{\left (-2 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{3}\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{2} \sqrt{-\frac{a^{7}}{d^{2}}} d \log \left (\frac{{\left (3 i \, \sqrt{2} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\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^{3}}\right ) + 3 \, \sqrt{2} \sqrt{-\frac{a^{7}}{d^{2}}} d \log \left (\frac{{\left (-3 i \, \sqrt{2} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\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^{3}}\right )}{2 \, 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]

1/2*(sqrt(2)*(-2*I*a^3*e^(2*I*d*x + 2*I*c) - 6*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3*sq
rt(2)*sqrt(-a^7/d^2)*d*log(1/3*(3*I*sqrt(2)*sqrt(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c) + 3*sqrt(2)*(a^3*e^(2*I*d*x +
2*I*c) + a^3)*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^3) + 3*sqrt(2)*sqrt(-
a^7/d^2)*d*log(1/3*(-3*I*sqrt(2)*sqrt(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c) + 3*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a
^3)*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^3))/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{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \cos \left (d x + c\right )^{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((I*a*tan(d*x + c) + a)^(7/2)*cos(d*x + c)^2, x)