3.108 \(\int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=201 \[ \frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[Out]

-1/2*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)-188/35*(a+I*a*tan(d*x+c))^(1/2)/a
/d+47/35*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/a/d-9/7*I*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3/a/d-tan(d*x+c)^
4/d/(a+I*a*tan(d*x+c))^(1/2)+223/105*(a+I*a*tan(d*x+c))^(3/2)/a^2/d

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Rubi [A]  time = 0.49, antiderivative size = 201, 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, 3597, 3592, 3527, 3480, 206} \[ \frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

-(ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(Sqrt[2]*Sqrt[a]*d)) - Tan[c + d*x]^4/(d*Sqrt[a + I*a*
Tan[c + d*x]]) - (188*Sqrt[a + I*a*Tan[c + d*x]])/(35*a*d) + (47*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(3
5*a*d) - (((9*I)/7)*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) + (223*(a + I*a*Tan[c + d*x])^(3/2))/(105
*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 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 3597

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

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-4 a+\frac {9}{2} i a \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}-\frac {2 \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {27 i a^2}{2}-\frac {47}{4} a^2 \tan (c+d x)\right ) \, dx}{7 a^3}\\ &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}-\frac {4 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {47 a^3}{2}-\frac {223}{8} i a^3 \tan (c+d x)\right ) \, dx}{35 a^4}\\ &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}-\frac {4 \int \sqrt {a+i a \tan (c+d x)} \left (\frac {223 i a^3}{8}+\frac {47}{2} a^3 \tan (c+d x)\right ) \, dx}{35 a^4}\\ &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {\tan ^4(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {188 \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {47 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {9 i \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 a d}+\frac {223 (a+i a \tan (c+d x))^{3/2}}{105 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.65, size = 123, normalized size = 0.61 \[ \frac {-\frac {840 e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-\left (\sec ^4(c+d x) (224 i \sin (2 (c+d x))+124 i \sin (4 (c+d x))+1484 \cos (2 (c+d x))+229 \cos (4 (c+d x))+1015)\right )}{840 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-840*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - Sec[c + d*x]^4*(1015 + 1484*C
os[2*(c + d*x)] + 229*Cos[4*(c + d*x)] + (224*I)*Sin[2*(c + d*x)] + (124*I)*Sin[4*(c + d*x)]))/(840*d*Sqrt[a +
I*a*Tan[c + d*x]])

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fricas [B]  time = 0.46, size = 389, normalized size = 1.94 \[ -\frac {105 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 105 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (353 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1708 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2030 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1260 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 105\right )}}{420 \, {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/420*(105*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(
I*d*x + I*c))*sqrt(1/(a*d^2))*log(4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/
(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 105*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*
I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d*e^(2*I*d*x + 2*I*c) + a*d
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 2*sqrt(2)*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1))*(353*e^(8*I*d*x + 8*I*c) + 1708*e^(6*I*d*x + 6*I*c) + 2030*e^(4*I*d*x + 4*I*c) + 12
60*e^(2*I*d*x + 2*I*c) + 105))/(a*d*e^(7*I*d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c
) + a*d*e^(I*d*x + I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 131, normalized size = 0.65 \[ \frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5}+\frac {8 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{3}-4 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}-\frac {a^{4}}{\sqrt {a +i a \tan \left (d x +c \right )}}}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^4*(1/7*(a+I*a*tan(d*x+c))^(7/2)-3/5*(a+I*a*tan(d*x+c))^(5/2)*a+4/3*(a+I*a*tan(d*x+c))^(3/2)*a^2-2*a^3*(a
+I*a*tan(d*x+c))^(1/2)-1/4*a^(7/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/2*a^4/(a+I*
a*tan(d*x+c))^(1/2))

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maxima [A]  time = 0.65, size = 156, normalized size = 0.78 \[ \frac {105 \, \sqrt {2} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 504 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 1120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 1680 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5} - \frac {420 \, a^{6}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{420 \, a^{6} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/420*(105*sqrt(2)*a^(11/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a))) + 120*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 504*(I*a*tan(d*x + c) + a)^(5/2)*a^3 + 1120*(I*a*ta
n(d*x + c) + a)^(3/2)*a^4 - 1680*sqrt(I*a*tan(d*x + c) + a)*a^5 - 420*a^6/sqrt(I*a*tan(d*x + c) + a))/(a^6*d)

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mupad [B]  time = 0.45, size = 144, normalized size = 0.72 \[ -\frac {1}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {4\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}+\frac {8\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^3\,d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a^4\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^2*d) - (4*(a + a*tan(c + d*x)*1i)^(1/2))/(a*d) - 1/(d*(a + a*tan(c + d*
x)*1i)^(1/2)) - (6*(a + a*tan(c + d*x)*1i)^(5/2))/(5*a^3*d) + (2*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a^4*d) + (2
^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(2*a^(1/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

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

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