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]

((-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)

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Rubi [A]  time = 0.338533, antiderivative size = 172, normalized size of antiderivative = 1., 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 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 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]

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)}{\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*}

Mathematica [A]  time = 1.632, 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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Maple [A]  time = 0.046, size = 113, normalized size = 0.7 \begin{align*}{\frac{2\,i}{d{a}^{3}} \left ({\frac{1}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{2\,a}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,{a}^{2}\sqrt{a+ia\tan \left ( dx+c \right ) }+{\frac{{a}^{3}}{2}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{\sqrt{2}}{4}{a}^{{\frac{5}{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+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)+1/2*a^3/
(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)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.23797, size = 1071, normalized size = 6.23 \begin{align*} \frac{\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 )} e^{\left (i \, d x + i \, c\right )} + \sqrt{2}{\left (-15 i \, a d e^{\left (6 i \, d x + 6 i \, c\right )} - 30 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - 15 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{1}{a d^{2}}} \log \left ({\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 (6 i \, d x + 6 i \, c\right )} + 30 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{1}{a d^{2}}} \log \left (-{\left (\sqrt{2} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{60 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

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)*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)*e^(I*d*x + I*c) + sqrt(2)*(-15*I*a*d*e^(6*I*d*x + 6*I*c) - 30*I*a*d*e^(4*I*d*x +
4*I*c) - 15*I*a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a*d^2))*log((sqrt(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c) +
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(
2)*(15*I*a*d*e^(6*I*d*x + 6*I*c) + 30*I*a*d*e^(4*I*d*x + 4*I*c) + 15*I*a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a*d^2)
)*log(-(sqrt(2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*
x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d
*e^(2*I*d*x + 2*I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

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(a*(I*tan(c + d*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{4}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

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)