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

Optimal. Leaf size=168 \[ \frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (((8*I)/35)*Sqrt[a + I*a*Tan[
c + d*x]])/d - (((2*I)/35)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d + (2*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c
+ d*x]])/(7*d) + (((62*I)/105)*(a + I*a*Tan[c + d*x])^(3/2))/(a*d)

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Rubi [A]  time = 0.347252, antiderivative size = 168, 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 = {3560, 3597, 3592, 3527, 3480, 206} \[ \frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (((8*I)/35)*Sqrt[a + I*a*Tan[
c + d*x]])/d - (((2*I)/35)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d + (2*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c
+ d*x]])/(7*d) + (((62*I)/105)*(a + I*a*Tan[c + d*x])^(3/2))/(a*d)

Rule 3560

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || 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 \tan ^4(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{2 \int \tan ^2(c+d x) \left (3 a+\frac{1}{2} i a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)} \, dx}{7 a}\\ &=-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}-\frac{4 \int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-i a^2+\frac{31}{4} a^2 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac{4 \int \sqrt{a+i a \tan (c+d x)} \left (-\frac{31 a^2}{4}-i a^2 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac{8 i \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}+\int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{8 i \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac{(2 i a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{35 d}-\frac{2 i \tan ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{35 d}+\frac{2 \tan ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{7 d}+\frac{62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}\\ \end{align*}

Mathematica [A]  time = 2.05825, size = 105, normalized size = 0.62 \[ \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{2}{105} \left (3 (5 \tan (c+d x)-i) \sec ^2(c+d x)-46 (\tan (c+d x)-i)\right )-i e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Tan[c + d*x]]*(((-I)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])/E^(I*(c + d*x)) + (
2*(-46*(-I + Tan[c + d*x]) + 3*Sec[c + d*x]^2*(-I + 5*Tan[c + d*x])))/105))/d

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Maple [A]  time = 0.054, size = 94, normalized size = 0.6 \begin{align*}{\frac{2\,i}{d{a}^{3}} \left ({\frac{1}{7} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{2\,a}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{a}^{2}}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{\sqrt{2}}{2}{a}^{{\frac{7}{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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^4,x)

[Out]

2*I/d/a^3*(1/7*(a+I*a*tan(d*x+c))^(7/2)-2/5*a*(a+I*a*tan(d*x+c))^(5/2)+2/3*(a+I*a*tan(d*x+c))^(3/2)*a^2-1/2*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((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.41636, size = 1031, normalized size = 6.14 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (368 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 448 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 560 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\right )} + 105 \, \sqrt{2}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (i \, \sqrt{2} d \sqrt{-\frac{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 (-2 i \, d x - 2 i \, c\right )}\right ) - 105 \, \sqrt{2}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (-i \, \sqrt{2} d \sqrt{-\frac{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 (-2 i \, d x - 2 i \, c\right )}\right )}{210 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(368*I*e^(6*I*d*x + 6*I*c) + 448*I*e^(4*I*d*x + 4*I*c) + 560*
I*e^(2*I*d*x + 2*I*c))*e^(I*d*x + I*c) + 105*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^
(2*I*d*x + 2*I*c) + d)*sqrt(-a/d^2)*log((I*sqrt(2)*d*sqrt(-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^(-2*I*d*x - 2*I*c)) - 105*sqrt(2)*(d*e^(6*I*d
*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-a/d^2)*log((-I*sqrt(2)*d*sqrt(-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^(-2*I*d*x - 2*I*c)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*tan(c + d*x)**4, x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out