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

Optimal. Leaf size=155 \[ \frac{B+i A}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{-B+i A}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{B+i A}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

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

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Rubi [A]  time = 0.129972, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3526, 3479, 3480, 206} \[ \frac{B+i A}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{-B+i A}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{B+i A}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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

Mathematica [A]  time = 2.46152, size = 176, normalized size = 1.14 \[ -\frac{e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt{1+e^{2 i (c+d x)}} \left (B \left (e^{2 i (c+d x)}-17 e^{4 i (c+d x)}+3\right )-i A \left (11 e^{2 i (c+d x)}+23 e^{4 i (c+d x)}+3\right )\right )+15 (B+i A) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(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))]*(B*(3 + E^((2*I)*(c + d*x)) - 17*E^((4*I)*(c
+ d*x))) - I*A*(3 + 11*E^((2*I)*(c + d*x)) + 23*E^((4*I)*(c + d*x)))) + 15*(I*A + B)*E^((5*I)*(c + d*x))*ArcSi
nh[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 = 0.023, size = 123, normalized size = 0.8 \begin{align*}{\frac{2\,i}{d} \left ( -{\frac{1}{5} \left ( -{\frac{A}{2}}-{\frac{i}{2}}B \right ) \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{-A+iB}{12\,a} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{-A+iB}{8\,{a}^{2}}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{ \left ( A-iB \right ) \sqrt{2}}{16}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}} \right ) } \end{align*}

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))^(5/2),x)

[Out]

2*I/d*(-1/5*(-1/2*A-1/2*I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/12/a*(-A+I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/8/a^2*(-A+I*B
)/(a+I*a*tan(d*x+c))^(1/2)-1/16*(A-I*B)/a^(5/2)*2^(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((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 = 1.98948, size = 1095, normalized size = 7.06 \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 (23 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (34 i \, A + 16 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (14 i \, A - 4 \, 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((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*s
qrt(-(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)) - 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)*((23*I*A + 17*B)*e^(6*I*d*x + 6*I*c) + (34*I*A + 16*B)*e^
(4*I*d*x + 4*I*c) + (14*I*A - 4*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((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{B \tan \left (d x + c\right ) + A}{{\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((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)/(I*a*tan(d*x + c) + a)^(5/2), x)