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

Optimal. Leaf size=167 \[ \frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(-7 B+i A) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {-11 B+5 i A}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

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

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Rubi [A]  time = 0.35, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3595, 3592, 3526, 3480, 206} \[ \frac {(-7 B+i A) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(-B+i A) \tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {-11 B+5 i A}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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 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 3592

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

Rule 3595

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

Rubi steps

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

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Mathematica [A]  time = 3.33, size = 167, normalized size = 1.00 \[ \frac {3 (A-i B) e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+A \left (7 e^{2 i (c+d x)}+8 e^{4 i (c+d x)}-1\right )+i B \left (13 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}-1\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(-1 + 7*E^((2*I)*(c + d*x)) + 8*E^((4*I)*(c + d*x))) + I*B*(-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))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])/(3*a*d*(1 + E
^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.64, size = 372, normalized size = 2.23 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {{\left (4 \, \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} - {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (8 i \, A - 38 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (7 i \, A - 13 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(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)*((8*I*A - 38*B)*e^(4*I*d*x + 4*I*c) + (7*I*A - 13*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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.23, size = 116, normalized size = 0.69 \[ -\frac {2 i \left (-i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}-\frac {a \left (5 i B +3 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{2} \left (i B +A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 137, normalized size = 0.82 \[ -\frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 48 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + 5 i \, B\right )} a^{2} - 2 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(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)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sq
rt(I*a*tan(d*x + c) + a))) - 48*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 4*(3*(I*a*tan(d*x + c) + a)*(3*A + 5*I*B)*a
^2 - 2*(A + I*B)*a^3)/(I*a*tan(d*x + c) + a)^(3/2))/(a^3*d)

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mupad [B]  time = 0.70, size = 186, normalized size = 1.11 \[ -\frac {\frac {A\,1{}\mathrm {i}}{3\,d}-\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {B\,a}{3}-\frac {5\,B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*a)/3 - (5*B*(a + a*tan(c + d*x)*1i))/2)/(a*d*(a + a*tan(c + d*x)*1i)^(3/2)) - ((A*1i)/(3*d) - (A*(a + a*ta
n(c + d*x)*1i)*3i)/(2*a*d))/(a + a*tan(c + d*x)*1i)^(3/2) - (2*B*(a + a*tan(c + d*x)*1i)^(1/2))/(a^2*d) + (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 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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