3.107 \(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {A-i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {A+3 i B}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

-1/8*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/4*(A-I*B)/a^2/d/(a+I*a*
tan(d*x+c))^(1/2)+1/5*(-A-I*B)/d/(a+I*a*tan(d*x+c))^(5/2)+1/6*(A+3*I*B)/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.23, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3590, 3526, 3479, 3480, 206} \[ \frac {A-i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {A+3 i B}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3590

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

Rubi steps

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

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Mathematica [A]  time = 3.23, size = 176, normalized size = 1.15 \[ \frac {e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \left (A \left (-e^{2 i (c+d x)}+17 e^{4 i (c+d x)}-3\right )-3 i B \left (-3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )\right )-15 (A-i B) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*((-3*I)*B*(1 - 3*E^((2*I)*(c + d*x)) + E^((4*I
)*(c + d*x))) + A*(-3 - E^((2*I)*(c + d*x)) + 17*E^((4*I)*(c + d*x)))) - 15*(A - I*B)*E^((5*I)*(c + d*x))*ArcS
inh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(240*a^2*d*E^((6*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.53, size = 394, normalized size = 2.58 \[ -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + 2 \, {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + 2 \, {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - \sqrt {2} {\left ({\left (17 \, A - 3 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (8 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (2 \, A - 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(1/2*(sqrt(2)*sqrt(1/2
)*(8*I*a^3*d*e^(2*I*d*x + 2*I*c) + 8*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^
5*d^2)) + 2*(4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*
A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(1/2*(sqrt(2)*sqrt(1/2)*(-8*I*a^3*d*e^(2*I*d*x + 2*I*c) - 8*I*a^3
*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + 2*(4*I*A + 4*B)*a*e^(I*d*x + I*c
))*e^(-I*d*x - I*c)/(I*A + B)) - sqrt(2)*((17*A - 3*I*B)*e^(6*I*d*x + 6*I*c) + 2*(8*A + 3*I*B)*e^(4*I*d*x + 4*
I*c) - 2*(2*A - 3*I*B)*e^(2*I*d*x + 2*I*c) - 3*A - 3*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-5*I*d*x - 5*I
*c)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 121, normalized size = 0.79 \[ \frac {-\frac {\left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}-\frac {2 \left (-\frac {3 i B}{4}-\frac {A}{4}\right )}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a \left (i B +A \right )}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {i B -A}{4 a \sqrt {a +i a \tan \left (d x +c \right )}}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*(-1/16*(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*(-3/4*I*B-1/4*A
)/(a+I*a*tan(d*x+c))^(3/2)-1/10*a*(A+I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/8/a*(-A+I*B)/(a+I*a*tan(d*x+c))^(1/2))

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maxima [A]  time = 0.75, size = 136, normalized size = 0.89 \[ \frac {\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (A - i \, B\right )} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A + 3 i \, B\right )} a - 12 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*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*(15*(I*a*tan(d*x + c) + a)^2*(A - I*B) + 10*(I*a*tan(d*x + c) + a)*(A + 3*I*B)*a
- 12*(A + I*B)*a^2)/(I*a*tan(d*x + c) + a)^(5/2))/(a^2*d)

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mupad [B]  time = 6.73, size = 186, normalized size = 1.22 \[ \frac {-\frac {A}{5}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6\,a}+\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {B\,1{}\mathrm {i}}{20\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}-\frac {\sqrt {2}\,A\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*(a + a*tan(c + d*x)*1i))/(6*a) - A/5 + (A*(a + a*tan(c + d*x)*1i)^2)/(4*a^2))/(d*(a + a*tan(c + d*x)*1i)^(
5/2)) + (B*1i)/(20*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (B*tan(c + d*x)^2*1i)/(4*d*(a + a*tan(c + d*x)*1i)^(5/2)
) - (2^(1/2)*B*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d) - (2^(1/2)*A*
atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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