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

Optimal. Leaf size=120 \[ -\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(1-i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3561, 3598, 12, 3544, 205} \[ -\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(1-i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*Sqrt[a + I*
a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - (((2*I)/3)*Sqrt[a + I*a*Tan[c + d*x]])/(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 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(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*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^
2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || 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 {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\left (\frac {i a}{2}-a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {3 a^2 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}+\frac {\left (2 i a^2\right ) \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 {(1-i) \sqrt {a} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 2.52, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.58, size = 355, normalized size = 2.96 \[ \frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{6 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(8*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^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3*I*c)) - 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*
I*a/d^2)*log((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) + d*sqrt(-2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 3*(d*e^(4*I*d*x + 4
*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I*a/d^2)*log((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) - d*sqrt(-2*I*a/d^2)*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.25, size = 196, normalized size = 1.63 \[ \frac {\left (3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a -4 i \tan \left (d x +c \right ) \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/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*t
an(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-4*I*tan(d*x+c)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-4*
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))*(a*(1+I*tan(d*x+c)))^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))
^(1/2)/(-I*a)^(1/2)

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maxima [B]  time = 0.67, size = 968, normalized size = 8.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(36*I + 36)*cos(3*d*x + 3*c)
- (12*I + 12)*cos(d*x + c) - (36*I - 36)*sin(3*d*x + 3*c) - (12*I - 12)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*
x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(36*I - 36)*cos(3*d*x + 3*c) - (12*I - 12)*cos(d*x + c) + (36*I + 36)*si
n(3*d*x + 3*c) + (12*I + 12)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a)
+ (((36*I + 36)*cos(2*d*x + 2*c)^2 + (36*I + 36)*sin(2*d*x + 2*c)^2 - (72*I + 72)*cos(2*d*x + 2*c) + 36*I + 36
)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x +
2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) +
1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + ((18*I - 18)*cos(2*d*x +
2*c)^2 + (18*I - 18)*sin(2*d*x + 2*c)^2 - (36*I - 36)*cos(2*d*x + 2*c) + 18*I - 18)*log(cos(d*x + c)^2 + sin(
d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x
+ 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x
+ 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((((12*I + 12)*cos(d*x + c) + (12*
I - 12)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I + 12)*cos(d*x + c) + (12*I - 12)*sin(d*x + c))*sin(2*d*x + 2
*c)^2 + (-(24*I + 24)*cos(d*x + c) - (24*I - 24)*sin(d*x + c))*cos(2*d*x + 2*c) + (12*I + 12)*cos(d*x + c) + (
12*I - 12)*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((12*I - 12)*cos(d*x + c
) - (12*I + 12)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I - 12)*cos(d*x + c) - (12*I + 12)*sin(d*x + c))*sin(2
*d*x + 2*c)^2 + (-(24*I - 24)*cos(d*x + c) + (24*I + 24)*sin(d*x + c))*cos(2*d*x + 2*c) + (12*I - 12)*cos(d*x
+ c) - (12*I + 12)*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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