\(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [311]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {3 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{4 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{16 d (a-i a \tan (c+d x))} \]

[Out]

-3/32*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)-1/4*I*a^4*(a+I*a*tan(d*x+c))^(
1/2)/d/(a-I*a*tan(d*x+c))^2-3/16*I*a^3*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3568, 44, 65, 212} \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {3 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{4 d (a-i a \tan (c+d x))^2}-\frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{16 d (a-i a \tan (c+d x))} \]

[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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {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 ) \text {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 ) \text {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 ) \text {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} \text {arctanh}\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 [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-1/4*I)*a^2*Hypergeometric2F1[1/2, 3, 3/2, (1 + I*Tan[c + d*x])/2]*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 845 vs. \(2 (110 ) = 220\).

Time = 123.88 (sec) , antiderivative size = 846, normalized size of antiderivative = 6.18

method	result	size
default	\(\text {Expression too large to display}\)	\(846\)

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16/d*(-tan(d*x+c)+I)^2*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*cos(d*x+c)^2*(-3*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+4*I*cos(d*x+c)^2*sin(d*x+c)+6
*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)+6*arc
tanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*(-cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)+6*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x
+c)+1))^(1/2))*cos(d*x+c)^2-6*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+
1))^(1/2))+6*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*
x+c)+6*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*cos(d*x+c)*sin(d*x+c)-6*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*
x+c)^2+6*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(
1/2))*cos(d*x+c)^3-3*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(
d*x+c)+1))^(1/2))+3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3
*I*cos(d*x+c)*sin(d*x+c)+4*cos(d*x+c)^3+3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1
))^(1/2))+7*cos(d*x+c)^2+3*cos(d*x+c))/(-I*cos(d*x+c)+sin(d*x+c)-I)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (102) = 204\).

Time = 0.25 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.92 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 3 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - \sqrt {2} {\left (-2 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{32 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(3*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(4*(a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(I*d*e^(2*I*
d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 3*sqrt(1/2)*sqrt(-a^5/d^2)*d*lo
g(4*(a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*
x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - sqrt(2)*(-2*I*a^2*e^(5*I*d*x + 5*I*c) - 7*I*a^2*e^(3*I*d*x + 3*I*c)
- 5*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 10 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}\right )}}{64 \, a d} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/64*I*(3*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 4*(3*(I*a*tan(d*x + c) + a)^(3/2)*a^4 - 10*sqrt(I*a*tan(d*x + c) + a)*a^5)/((I*a*tan(d*x + c
) + a)^2 - 4*(I*a*tan(d*x + c) + a)*a + 4*a^2))/(a*d)

Giac [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2), x)