3.188 \(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=194 \[ \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

(1/4+1/4*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(3/2)/d+1/6*(11*A+5*I*B
)/a/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-1/6*(25*A+7*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(1/
2)+1/3*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.58, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ -\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((1/4 + I/4)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) +
(A + I*B)/(3*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (11*A + (5*I)*B)/(6*a*d*Sqrt[Tan[c + d*x]]*
Sqrt[a + I*a*Tan[c + d*x]]) - ((25*A + (7*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*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 3596

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*A + b*B)*(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*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

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 {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (7 A+i B)-2 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{4} a^2 (25 A+7 i B)-\frac {1}{2} a^2 (11 i A-5 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {2 \int \frac {3 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^5}\\ &=\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i A+B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {(A-i B) \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 )}{2 d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11 A+5 i B}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(25 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 5.42, size = 237, normalized size = 1.22 \[ \frac {i e^{-2 i (c+d x)} \sqrt {\tan (c+d x)} \csc (c+d x) \sec (c+d x) \left (\sqrt {-1+e^{2 i (c+d x)}} \left (A \left (-13 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}-1\right )+i B \left (-7 e^{2 i (c+d x)}+8 e^{4 i (c+d x)}-1\right )\right )-3 (A-i B) e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{12 a d \sqrt {-1+e^{2 i (c+d x)}} (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)),x]

[Out]

((I/12)*(Sqrt[-1 + E^((2*I)*(c + d*x))]*(I*B*(-1 - 7*E^((2*I)*(c + d*x)) + 8*E^((4*I)*(c + d*x))) + A*(-1 - 13
*E^((2*I)*(c + d*x)) + 38*E^((4*I)*(c + d*x)))) - 3*(A - I*B)*E^((3*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*A
rcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Csc[c + d*x]*Sec[c + d*x]*Sqrt[Tan[c + d*x]])/(a*d*E^(
(2*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

fricas [B]  time = 1.43, size = 509, normalized size = 2.62 \[ \frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (-38 i \, A + 8 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-25 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (14 i \, A - 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(1/2)*(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^
2))*log((2*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I
*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 3*sqrt(1/2)*(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt((I*A^2 +
2*A*B - I*B^2)/(a^3*d^2))*log(-(2*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))*e^(I*d*x + I*c) - sq
rt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + sqrt(2)*((-38*I*A + 8*B)*e^(6*I*d*x + 6*I*c) + (-25*I*A + B
)*e^(4*I*d*x + 4*I*c) + (14*I*A - 8*B)*e^(2*I*d*x + 2*I*c) + I*A - B)*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)))/(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d*e^(3*I*d*x + 3*I*c)
)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/((I*a*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^(3/2)), x)

________________________________________________________________________________________

maple [B]  time = 0.30, size = 931, normalized size = 4.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/24/d*(a*(1+I*tan(d*x+c)))^(1/2)*(3*I*B*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+9*I*A*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-3*A*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-9*I*B*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+28*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+9*B*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-3*I*A*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))*t
an(d*x+c)*a-256*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+9*A*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+100*A*
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3-36*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(
-I*a)^(1/2)*tan(d*x+c)-3*B*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)*a+64*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+
48*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)-204*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)*tan(d*x+c))/a^2/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^3

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(3/2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________