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

Optimal. Leaf size=121 \[ -\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {-B+i A}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {B+i A}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-1/4*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+1/2*(I*A+B)/a/d/(a+I*a*ta
n(d*x+c))^(1/2)+1/3*(I*A-B)/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3526, 3479, 3480, 206} \[ -\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {-B+i A}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {B+i A}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

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

Rule 3526

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

Rubi steps

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

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Mathematica [A]  time = 2.59, size = 143, normalized size = 1.18 \[ \frac {\sqrt {1+e^{2 i (c+d x)}} \left (4 A e^{2 i (c+d x)}+A-i B \left (-1+2 e^{2 i (c+d x)}\right )\right )-3 (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^{3/2} (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + E^((2*I)*(c + d*x))]*(A + 4*A*E^((2*I)*(c + d*x)) - I*B*(-1 + 2*E^((2*I)*(c + d*x)))) - 3*(A - I*B)*
E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(3*a*d*(1 + E^((2*I)*(c + d*x)))^(3/2)*(-I + Tan[c + d*x])*Sqrt[
a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.49, size = 373, normalized size = 3.08 \[ -\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} - {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (4 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (5 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sqrt(1/2)*a^2*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((4*sqrt(2)*sqrt(1/2)*(
a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2)) +
(4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(
a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-(4*sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^3*d^2)) - (4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*
A + B)) - sqrt(2)*((4*I*A + 2*B)*e^(4*I*d*x + 4*I*c) + (5*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)))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(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), x)

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maple [A]  time = 0.17, size = 96, normalized size = 0.79 \[ \frac {2 i \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}-\frac {-\frac {A}{2}-\frac {i B}{2}}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/d*(-1/8*(A-I*B)/a^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/3*(-1/2*A-1/2*I*B)
/(a+I*a*tan(d*x+c))^(3/2)-1/4/a*(-A+I*B)/(a+I*a*tan(d*x+c))^(1/2))

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maxima [A]  time = 0.99, size = 113, normalized size = 0.93 \[ \frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A - i \, B\right )} + 2 \, {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*I*(3*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a)))/sqrt(a) + 4*(3*(I*a*tan(d*x + c) + a)*(A - I*B) + 2*(A + I*B)*a)/(I*a*tan(d*x + c) + a)^(3/2
))/(a*d)

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mupad [B]  time = 6.99, size = 162, normalized size = 1.34 \[ \frac {\frac {A\,1{}\mathrm {i}}{3\,d}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {B}{3}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,a}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*1i)/(3*d) + (A*(a + a*tan(c + d*x)*1i)*1i)/(2*a*d))/(a + a*tan(c + d*x)*1i)^(3/2) - (B/3 - (B*(a + a*tan(c
+ d*x)*1i))/(2*a))/(d*(a + a*tan(c + d*x)*1i)^(3/2)) - (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
)/(2*(-a)^(1/2)))*1i)/(4*(-a)^(3/2)*d) - (2^(1/2)*B*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2)))
)/(4*a^(3/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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