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

Optimal. Leaf size=205 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

-1/8*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)-89/5*(a+I*a*tan(d*x+c))^(1/2)/a^3
/d+89/20*tan(d*x+c)^2/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1/5*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^(5/2)+7/10*I*tan(d*
x+c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)+361/60*(a+I*a*tan(d*x+c))^(3/2)/a^4/d

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Rubi [A]  time = 0.52, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3558, 3595, 3592, 3527, 3480, 206} \[ \frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - Tan[c + d*x]^4/(5*d*(a + I*a*Ta
n[c + d*x])^(5/2)) + (((7*I)/10)*Tan[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (89*Tan[c + d*x]^2)/(20*
a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (89*Sqrt[a + I*a*Tan[c + d*x]])/(5*a^3*d) + (361*(a + I*a*Tan[c + d*x])^(3
/2))/(60*a^4*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 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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c -
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

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 ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\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+\frac {13}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan ^2(c+d x) \left (-\frac {63 i a^2}{2}-\frac {141}{4} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \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 {267 a^3}{2}-\frac {1083}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {361 (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 {1083 i a^3}{8}+\frac {267}{2} a^3 \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}-\frac {\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 {\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 {\tan ^4(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 i \tan ^3(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \tan ^2(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {89 \sqrt {a+i a \tan (c+d x)}}{5 a^3 d}+\frac {361 (a+i a \tan (c+d x))^{3/2}}{60 a^4 d}\\ \end {align*}

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Mathematica [A]  time = 1.94, size = 149, normalized size = 0.73 \[ -\frac {e^{-4 i (c+d x)} \left (-33 e^{2 i (c+d x)}+348 e^{4 i (c+d x)}+1527 e^{6 i (c+d x)}+983 e^{8 i (c+d x)}+15 e^{5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sinh ^{-1}\left (e^{i (c+d x)}\right )+3\right )}{60 a^2 d \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(3 - 33*E^((2*I)*(c + d*x)) + 348*E^((4*I)*(c + d*x)) + 1527*E^((6*I)*(c + d*x)) + 983*E^((8*I)*(c + d*x
)) + 15*E^((5*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(3/2)*ArcSinh[E^(I*(c + d*x))])/(a^2*d*E^((4*I)*(c + d*x
))*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.46, size = 342, normalized size = 1.67 \[ -\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (983 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1527 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 348 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 33 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(4*(sqrt(2)*
sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x
+ I*c))*e^(-I*d*x - I*c)) - 15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*
d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(
a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(983*e^(8*I*d*x +
8*I*c) + 1527*e^(6*I*d*x + 6*I*c) + 348*e^(4*I*d*x + 4*I*c) - 33*e^(2*I*d*x + 2*I*c) + 3))/(a^3*d*e^(7*I*d*x
+ 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 131, normalized size = 0.64 \[ \frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {31 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {3 a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{4}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^4*(1/3*(a+I*a*tan(d*x+c))^(3/2)-3*a*(a+I*a*tan(d*x+c))^(1/2)-31/8*a^2/(a+I*a*tan(d*x+c))^(1/2)+3/4*a^3/(
a+I*a*tan(d*x+c))^(3/2)-1/10*a^4/(a+I*a*tan(d*x+c))^(5/2)-1/16*a^(3/2)*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 [A]  time = 0.67, size = 157, normalized size = 0.77 \[ \frac {15 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 160 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 1440 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {12 \, {\left (155 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} - 30 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{5} + 4 \, a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{6} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 160*(I*a*tan(d*x + c) + a)^(3/2)*a^2 - 1440*sqrt(I*a*tan(d*x + c) + a)*a^3 - 12*(155*(I*a*ta
n(d*x + c) + a)^2*a^4 - 30*(I*a*tan(d*x + c) + a)*a^5 + 4*a^6)/(I*a*tan(d*x + c) + a)^(5/2))/(a^6*d)

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mupad [B]  time = 3.91, size = 140, normalized size = 0.68 \[ -\frac {6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^4\,d}-\frac {\frac {31\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {3\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}+\frac {a^2}{5}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^4*d) - (6*(a + a*tan(c + d*x)*1i)^(1/2))/(a^3*d) - ((31*(a + a*tan(c +
d*x)*1i)^2)/4 - (3*a*(a + a*tan(c + d*x)*1i))/2 + a^2/5)/(a^2*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*atan
((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(8*a^(5/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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