3.104 \(\int \frac{\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=255 \[ -\frac{(-89 B+39 i A) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(-361 B+151 i A) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac{(-89 B+39 i A) \sqrt{a+i a \tan (c+d x)}}{5 a^3 d}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(-B+i A) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[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]^4)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*A + (21*I)*B)*Tan[c + d*x]^3)/(30*a*d*(a + I*a*Tan[c + d*x])
^(3/2)) - (((39*I)*A - 89*B)*Tan[c + d*x]^2)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((39*I)*A - 89*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/(5*a^3*d) - (((151*I)*A - 361*B)*(a + I*a*Tan[c + d*x])^(3/2))/(60*a^4*d)

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Rubi [A]  time = 0.779281, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3595, 3592, 3527, 3480, 206} \[ -\frac{(-89 B+39 i A) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(-361 B+151 i A) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac{(-89 B+39 i A) \sqrt{a+i a \tan (c+d x)}}{5 a^3 d}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{(-B+i A) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^4*(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]^4)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*A + (21*I)*B)*Tan[c + d*x]^3)/(30*a*d*(a + I*a*Tan[c + d*x])
^(3/2)) - (((39*I)*A - 89*B)*Tan[c + d*x]^2)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((39*I)*A - 89*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/(5*a^3*d) - (((151*I)*A - 361*B)*(a + I*a*Tan[c + d*x])^(3/2))/(60*a^4*d)

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 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 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, 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 ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\tan ^3(c+d x) \left (4 a (i A-B)+\frac{1}{2} a (3 A+13 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 ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\tan ^2(c+d x) \left (-\frac{3}{2} a^2 (11 A+21 i B)+\frac{3}{4} a^2 (17 i A-47 B) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{2} a^3 (39 i A-89 B)-\frac{3}{8} a^3 (151 A+361 i B) \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac{\int \sqrt{a+i a \tan (c+d x)} \left (\frac{3}{8} a^3 (151 A+361 i B)-\frac{3}{2} a^3 (39 i A-89 B) \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(39 i A-89 B) \sqrt{a+i a \tan (c+d x)}}{5 a^3 d}-\frac{(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}+\frac{(A-i B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac{(i A-B) \tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(39 i A-89 B) \sqrt{a+i a \tan (c+d x)}}{5 a^3 d}-\frac{(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 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 )}{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 ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(11 A+21 i B) \tan ^3(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac{(39 i A-89 B) \tan ^2(c+d x)}{20 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(39 i A-89 B) \sqrt{a+i a \tan (c+d x)}}{5 a^3 d}-\frac{(151 i A-361 B) (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}\\ \end{align*}

Mathematica [A]  time = 5.30489, size = 191, normalized size = 0.75 \[ \frac{\frac{120 (B+i A) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{5/2}}+\sec ^4(c+d x) ((747 B-317 i A) \cos (2 (c+d x))+(493 B-233 i A) \cos (4 (c+d x))+340 A \sin (2 (c+d x))+230 A \sin (4 (c+d x))-84 i A+780 i B \sin (2 (c+d x))+490 i B \sin (4 (c+d x))+174 B)}{120 a^2 d (\tan (c+d x)-i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((120*(I*A + B)*E^((5*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(5/2) + Sec[c + d*x]^4
*((-84*I)*A + 174*B + ((-317*I)*A + 747*B)*Cos[2*(c + d*x)] + ((-233*I)*A + 493*B)*Cos[4*(c + d*x)] + 340*A*Si
n[2*(c + d*x)] + (780*I)*B*Sin[2*(c + d*x)] + 230*A*Sin[4*(c + d*x)] + (490*I)*B*Sin[4*(c + d*x)]))/(120*a^2*d
*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.033, size = 181, normalized size = 0.7 \begin{align*}{\frac{2\,i}{{a}^{4}d} \left ( -{\frac{i}{3}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}+3\,iBa\sqrt{a+ia\tan \left ( dx+c \right ) }+A\sqrt{a+ia\tan \left ( dx+c \right ) }a+{\frac{{a}^{2} \left ( 31\,iB+17\,A \right ) }{8}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{{a}^{3} \left ( 9\,iB+7\,A \right ) }{12} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{4} \left ( A+iB \right ) }{10} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{ \left ( A-iB \right ) \sqrt{2}}{16}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2*I/d/a^4*(-1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)+3*I*B*a*(a+I*a*tan(d*x+c))^(1/2)+A*(a+I*a*tan(d*x+c))^(1/2)*a+1/8
*a^2*(31*I*B+17*A)/(a+I*a*tan(d*x+c))^(1/2)-1/12*a^3*(9*I*B+7*A)/(a+I*a*tan(d*x+c))^(3/2)+1/10*a^4*(A+I*B)/(a+
I*a*tan(d*x+c))^(5/2)-1/16*a^(3/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

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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)^4*(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 = 2.16433, size = 1274, normalized size = 5. \begin{align*} \frac{\sqrt{2}{\left ({\left (463 i \, A - 983 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (657 i \, A - 1527 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (168 i \, A - 348 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-23 i \, A + 33 \, 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 )} - 15 \, \sqrt{\frac{1}{2}}{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \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}}{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt{-\frac{A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} \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 )}{120 \,{\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(sqrt(2)*((463*I*A - 983*B)*e^(8*I*d*x + 8*I*c) + (657*I*A - 1527*B)*e^(6*I*d*x + 6*I*c) + (168*I*A - 34
8*B)*e^(4*I*d*x + 4*I*c) + (-23*I*A + 33*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) - 15*sqrt(1/2)*(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt(-(A^2 - 2*I*A*B
- B^2)/(a^5*d^2))*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)) + 15*sqrt(1/2)*(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)
/(a^5*d^2))*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)))/(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))

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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)**4*(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 )^{4}}{{\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)^4*(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)^4/(I*a*tan(d*x + c) + a)^(5/2), x)