3.109 \(\int \frac {\tan ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=172 \[ -\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[Out]

-1/2*I*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)+28/5*I*(a+I*a*tan(d*x+c))^(1/2)
/a/d-7/5*I*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/a/d-tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-23/15*I*(a+I*a*ta
n(d*x+c))^(3/2)/a^2/d

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Rubi [A]  time = 0.34, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) - Tan[c + d*x]^3/(d*Sqrt[a +
I*a*Tan[c + d*x]]) + (((28*I)/5)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (((7*I)/5)*Tan[c + d*x]^2*Sqrt[a + I*a*Ta
n[c + d*x]])/(a*d) - (((23*I)/15)*(a + I*a*Tan[c + d*x])^(3/2))/(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 ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-3 a+\frac {7}{2} i a \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {2 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-7 i a^2-\frac {23}{4} a^2 \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac {2 \int \sqrt {a+i a \tan (c+d x)} \left (\frac {23 a^2}{4}-7 i a^2 \tan (c+d x)\right ) \, dx}{5 a^3}\\ &=-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}-\frac {i \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {\tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {28 i \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {7 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {23 i (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 122, normalized size = 0.71 \[ \frac {i \sec ^3(c+d x) (20 i \sin (c+d x)+44 i \sin (3 (c+d x))+185 \cos (c+d x)+59 \cos (3 (c+d x)))-\frac {60 i e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}}{60 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-60*I)*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + I*Sec[c + d*x]^3*(185*Cos[
c + d*x] + 59*Cos[3*(c + d*x)] + (20*I)*Sin[c + d*x] + (44*I)*Sin[3*(c + d*x)]))/(60*d*Sqrt[a + I*a*Tan[c + d*
x]])

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fricas [B]  time = 0.46, size = 340, normalized size = 1.98 \[ \frac {\sqrt {2} {\left (-15 i \, a d e^{\left (5 i \, d x + 5 i \, c\right )} - 30 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} {\left (15 i \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 30 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + 15 i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (206 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 410 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 330 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 30 i\right )}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/60*(sqrt(2)*(-15*I*a*d*e^(5*I*d*x + 5*I*c) - 30*I*a*d*e^(3*I*d*x + 3*I*c) - 15*I*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)) + sqrt(2)*(15*I*a*d*e^(5*I*d*x + 5*I*c) + 30*I*a*d*e^(3*I*d*x + 3*I*c) + 15*I*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))*sqr
t(1/(a*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))*(206*I*e^(6*I*
d*x + 6*I*c) + 410*I*e^(4*I*d*x + 4*I*c) + 330*I*e^(2*I*d*x + 2*I*c) + 30*I))/(a*d*e^(5*I*d*x + 5*I*c) + 2*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 \[ \int \frac {\tan \left (d x + c\right )^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^4/sqrt(I*a*tan(d*x + c) + a), x)

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maple [A]  time = 0.18, size = 113, normalized size = 0.66 \[ \frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a}{3}+2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

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

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maxima [A]  time = 0.74, size = 138, normalized size = 0.80 \[ \frac {i \, {\left (15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 80 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 240 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4} + \frac {60 \, a^{5}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}}{60 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/60*I*(15*sqrt(2)*a^(9/2)*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))) + 24*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 80*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 240*sqrt(I*a*ta
n(d*x + c) + a)*a^4 + 60*a^5/sqrt(I*a*tan(d*x + c) + a))/(a^5*d)

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mupad [B]  time = 4.18, size = 129, normalized size = 0.75 \[ \frac {1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a^3\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

1i/(d*(a + a*tan(c + d*x)*1i)^(1/2)) + ((a + a*tan(c + d*x)*1i)^(1/2)*4i)/(a*d) - ((a + a*tan(c + d*x)*1i)^(3/
2)*4i)/(3*a^2*d) + ((a + a*tan(c + d*x)*1i)^(5/2)*2i)/(5*a^3*d) + (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1
i)^(1/2))/(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 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**4/sqrt(I*a*(tan(c + d*x) - I)), x)

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