\(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [324]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{8 d (a-i a \tan (c+d x))} \]

[Out]

1/16*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)-1/2*I*a^5*(a+I*a*tan(d*x+c))^(1
/2)/d/(a-I*a*tan(d*x+c))^2+1/8*I*a^4*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 43, 44, 65, 212} \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{8 d (a-i a \tan (c+d x))} \]

[In]

Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((I/8)*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) - ((I/2)*a^5*Sqrt[a + I*a*Ta
n[c + d*x]])/(d*(a - I*a*Tan[c + d*x])^2) + ((I/8)*a^4*Sqrt[a + I*a*Tan[c + d*x]])/(d*(a - I*a*Tan[c + d*x]))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 {\sqrt {a+x}}{(a-x)^3} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 d} \\ & = -\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{8 d (a-i a \tan (c+d x))}+\frac {\left (i a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{16 d} \\ & = -\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{8 d (a-i a \tan (c+d x))}+\frac {\left (i a^4\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 d} \\ & = \frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^4 \sqrt {a+i a \tan (c+d x)}}{8 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 = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{3/2}}{12 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-1/12*I)*a^2*Hypergeometric2F1[3/2, 3, 5/2, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^(3/2))/d

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 868 vs. \(2 (110 ) = 220\).

Time = 123.98 (sec) , antiderivative size = 869, normalized size of antiderivative = 6.34

method	result	size
default	\(\text {Expression too large to display}\)	\(869\)

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*(-tan(d*x+c)+I)^3*(a*(1+I*tan(d*x+c)))^(1/2)*a^3*cos(d*x+c)^3*(I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct
an((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3*I*cos(d*x+c)^2+2*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct
anh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*x+c)-4*I*cos(d*x+c)^3-2*(-c
os(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)^3-2*(-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)+I
*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*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*sin(d*x+c)-2*arctan
h(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)^
2-2*(-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)-2*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+I*cos(d*x+c)+(-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*
cos(d*x+c)^2*sin(d*x+c)-2*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)^3+(-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))+sin(d*x+c)*cos(d*x+c))/(-2*I*cos(d*x+c)^2+2*sin(d*x+c)*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. 261 vs. \(2 (102) = 204\).

Time = 0.24 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.91 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {\sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) + \sqrt {2} {\left (-2 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 3 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/16*(sqrt(1/2)*sqrt(-a^7/d^2)*d*log(4*(a^4*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^7/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^3) - sqrt(1/2)*sqrt(-a^7/d^2)*d*log(4*(
a^4*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^7/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^3) + sqrt(2)*(-2*I*a^3*e^(5*I*d*x + 5*I*c) - 3*I*a^3*e^(3*I*d*x + 3*I*c) - I*a
^3*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i \, {\left (\sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 2 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}\right )}}{32 \, a d} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-1/32*I*(sqrt(2)*a^(9/2)*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*((I*a*tan(d*x + c) + a)^(3/2)*a^5 + 2*sqrt(I*a*tan(d*x + c) + a)*a^6)/((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. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \]

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2), x)