3.216 \(\int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=161 \[ -\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

(-1/2+1/2*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d/a^(1/2)+7/3*I*(a+I*a*tan(d*x+c
))^(1/2)/a/d/tan(d*x+c)^(1/2)+1/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2)-5/3*(a+I*a*tan(d*x+c))^(1/2)/a/d/t
an(d*x+c)^(3/2)

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Rubi [A]  time = 0.41, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3559, 3598, 12, 3544, 205} \[ -\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + 1/(d*Tan
[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (5*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + (((7
*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {5 a}{2}-2 i a \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {7 i a^2}{4}-\frac {5}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {3 a^3 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {(i a) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {5 \sqrt {a+i a \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {7 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.85, size = 159, normalized size = 0.99 \[ \frac {i \left (-18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+3\right )}{3 \sqrt {2} d \left (-1+e^{4 i (c+d x)}\right ) \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((I/3)*(3 - 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]*(-1 + E^((4*I)*(c + d*x)))*Sqrt[Tan[c + d*x]])

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fricas [B]  time = 0.52, size = 403, normalized size = 2.50 \[ -\frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (7 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 11 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 15 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} + 3 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (\frac {1}{4} \, a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 3 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (-\frac {1}{4} \, a d \sqrt {-\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )}{12 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)
)*(7*e^(6*I*d*x + 6*I*c) - 11*e^(4*I*d*x + 4*I*c) - 15*e^(2*I*d*x + 2*I*c) + 3) + 3*(a*d*e^(5*I*d*x + 5*I*c) -
2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-2*I/(a*d^2))*log(1/4*a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x +
I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) +
1))*(e^(2*I*d*x + 2*I*c) + 1)) - 3*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c)
)*sqrt(-2*I/(a*d^2))*log(-1/4*a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)))/(a*d*e^(5*I*d*
x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(I*a*tan(d*x + c) + a)*tan(d*x + c)^(5/2)), x)

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maple [B]  time = 0.24, size = 398, normalized size = 2.47 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a -3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +6 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a +36 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+28 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )+8 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{12 d a \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(3/2)*(3*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-3*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+6*2^(1/2)*ln((2*2^(1
/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+36*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+28*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*
a)^(1/2)*tan(d*x+c)^3+8*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(I*a*tan(d*x + c) + a)*tan(d*x + c)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)

[Out]

Integral(1/(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**(5/2)), x)

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