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]

-(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)

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Rubi [A]  time = 0.488413, antiderivative size = 201, normalized size of antiderivative = 1., 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 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 ^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*}

Mathematica [A]  time = 1.54846, size = 123, normalized size = 0.61 \[ \frac{\sec ^4(c+d x) (-(224 i \sin (2 (c+d x))+124 i \sin (4 (c+d x))+1484 \cos (2 (c+d x))+229 \cos (4 (c+d x))+1015))-\frac{840 e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}}{840 d \sqrt{a+i a \tan (c+d x)}} \]

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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Maple [A]  time = 0.045, size = 131, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{d{a}^{4}} \left ( 1/7\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{7/2}-3/5\,a \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{5/2}+4/3\, \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3/2}{a}^{2}-2\,{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-1/2\,{\frac{{a}^{4}}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}-1/4\,{a}^{7/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.38456, size = 1193, normalized size = 5.94 \begin{align*} -\frac{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 )} e^{\left (i \, d x + i \, c\right )} + 105 \, \sqrt{2}{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, 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 )} \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 ) - 105 \, \sqrt{2}{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, 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 )} \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 )}{420 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, 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)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/420*(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) + 1260*e^(2*I*d*x + 2*I*c) + 105)*e^(I*d*x + I*c) + 105*sqrt(2)*(a*d*e^(8*I*d*x + 8*I*c)
+ 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) + 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)) - 105*sqrt(2)*(a*d*e^(8*I*d*x + 8*I*c) + 3*a*d*e^(6*I*d*x + 6*I*c) + 3*a*
d*e^(4*I*d*x + 4*I*c) + 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^(8*I*d*x + 8*I*c) + 3*a*d*e^(6*I*d*x + 6*I*c) + 3*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 ^{5}{\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)**5/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**5/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 )^{5}}{\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)^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)