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

Optimal. Leaf size=167 \[ -\frac{-31 B+i A}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{-13 B+3 i A}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((I*A - B)*Tan[c + d
*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((3*I)*A - 13*B)/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (I*A - 31
*B)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.39508, antiderivative size = 167, normalized size of antiderivative = 1., 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, 3590, 3526, 3480, 206} \[ -\frac{-31 B+i A}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{-13 B+3 i A}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((I*A - B)*Tan[c + d
*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((3*I)*A - 13*B)/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (I*A - 31
*B)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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]

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\tan (c+d x) \left (2 a (i A-B)-\frac{1}{2} a (A-9 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{i \int \frac{\frac{1}{2} a^2 (3 i A-13 B)-a^2 (A-9 i B) \tan (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{10 a^4}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i A-31 B}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(A-i B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i A-31 B}{20 a^2 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 )}{4 a^2 d}\\ &=\frac{(i A+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{(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{i A-31 B}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.15901, size = 176, normalized size = 1.05 \[ \frac{e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt{1+e^{2 i (c+d x)}} \left (B \left (-19 e^{2 i (c+d x)}+83 e^{4 i (c+d x)}+3\right )-3 i A \left (-3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )\right )+15 (B+i A) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^2*(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)*A*(1 - 3*E^((2*I)*(c + d*x)) + E^((4*I
)*(c + d*x))) + B*(3 - 19*E^((2*I)*(c + d*x)) + 83*E^((4*I)*(c + d*x)))) + 15*(I*A + B)*E^((5*I)*(c + d*x))*Ar
cSinh[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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Maple [A]  time = 0.031, size = 124, normalized size = 0.7 \begin{align*}{\frac{-2\,i}{{a}^{2}d} \left ( -{ \left ( -{\frac{7\,i}{8}}B-{\frac{A}{8}} \right ){\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{a \left ( 3\,A+5\,iB \right ) }{12} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{2} \left ( A+iB \right ) }{10} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{\sqrt{2}}{2} \left ({\frac{A}{8}}-{\frac{i}{8}}B \right ){\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}} \right ) } \end{align*}

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))^(5/2),x)

[Out]

-2*I/d/a^2*(-(-7/8*I*B-1/8*A)/(a+I*a*tan(d*x+c))^(1/2)-1/12*a*(3*A+5*I*B)/(a+I*a*tan(d*x+c))^(3/2)+1/10*a^2*(A
+I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/2*(1/8*A-1/8*I*B)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2
)/a^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.02363, size = 1092, normalized size = 6.54 \begin{align*} \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 (6 i \, d x + 6 i \, c\right )} \log \left (\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\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 (6 i \, d x + 6 i \, c\right )} \log \left (-\frac{{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 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}} 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 (-3 i \, A + 83 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (6 i \, A + 64 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (6 i \, A - 16 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \end{align*}

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))^(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^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a^3*d*sq
rt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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^(6*I*d*x + 6*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))
*e^(2*I*d*x + 2*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))*e^(
I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) + sqrt(2)*((-3*I*A + 83*B)*e^(6*I*d*x + 6*I*c) + (6*I*A + 64*B)*e^(4
*I*d*x + 4*I*c) + (6*I*A - 16*B)*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x
+ I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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))**(5/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

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