3.118 \(\int \frac{\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}-\frac{10 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

((-I/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^3/(3*d*(a +
I*a*Tan[c + d*x])^(3/2)) + (((5*I)/2)*Tan[c + d*x]^2)/(a*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((10*I)*Sqrt[a + I*a*
Tan[c + d*x]])/(a^2*d) + (((7*I)/2)*(a + I*a*Tan[c + d*x])^(3/2))/(a^3*d)

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Rubi [A]  time = 0.353426, antiderivative size = 174, 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, 3595, 3592, 3527, 3480, 206} \[ \frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}-\frac{10 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-I/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^3/(3*d*(a +
I*a*Tan[c + d*x])^(3/2)) + (((5*I)/2)*Tan[c + d*x]^2)/(a*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((10*I)*Sqrt[a + I*a*
Tan[c + d*x]])/(a^2*d) + (((7*I)/2)*(a + I*a*Tan[c + d*x])^(3/2))/(a^3*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 3595

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(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 - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 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)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac{\int \frac{\tan ^2(c+d x) \left (-3 a+\frac{9}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \tan (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-15 i a^2-\frac{63}{4} a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}+\frac{\int \sqrt{a+i a \tan (c+d x)} \left (\frac{63 a^2}{4}-15 i a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{10 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}+\frac{\int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{10 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}-\frac{i \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2 a d}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i \tan ^2(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{10 i \sqrt{a+i a \tan (c+d x)}}{a^2 d}+\frac{7 i (a+i a \tan (c+d x))^{3/2}}{2 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.18564, size = 165, normalized size = 0.95 \[ -\frac{i e^{-4 i (c+d x)} \sec ^2(c+d x) \left (\left (18 e^{2 i (c+d x)}+87 e^{4 i (c+d x)}+52 e^{6 i (c+d x)}-1\right ) \sqrt{1+e^{2 i (c+d x)}}+3 e^{3 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{24 a d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-I/24)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-1 + 18*E^((2*I)*(c + d*x)) + 87*E^((4*I)*(c + d*x)) + 52*E^((6*I)*(c
+ d*x))) + 3*E^((3*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^2*ArcSinh[E^(I*(c + d*x))])*Sec[c + d*x]^2)/(a*d*E
^((4*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.022, size = 113, normalized size = 0.7 \begin{align*}{\frac{2\,i}{d{a}^{3}} \left ({\frac{1}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-2\,a\sqrt{a+ia\tan \left ( dx+c \right ) }-{\frac{\sqrt{2}}{8}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) }-{\frac{7\,{a}^{2}}{4}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{{a}^{3}}{6} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \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))^(3/2),x)

[Out]

2*I/d/a^3*(1/3*(a+I*a*tan(d*x+c))^(3/2)-2*a*(a+I*a*tan(d*x+c))^(1/2)-1/8*a^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*ta
n(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-7/4/(a+I*a*tan(d*x+c))^(1/2)*a^2+1/6*a^3/(a+I*a*tan(d*x+c))^(3/2))

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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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.24789, size = 986, normalized size = 5.67 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-52 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 87 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d x + i \, c\right )} + \sqrt{\frac{1}{2}}{\left (-3 i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{1}{a^{3} d^{2}}} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} 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{\frac{1}{2}}{\left (3 i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt{\frac{1}{a^{3} d^{2}}} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} 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 )}{12 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 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))^(3/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-52*I*e^(6*I*d*x + 6*I*c) - 87*I*e^(4*I*d*x + 4*I*c) - 18*I*e
^(2*I*d*x + 2*I*c) + I)*e^(I*d*x + I*c) + sqrt(1/2)*(-3*I*a^2*d*e^(6*I*d*x + 6*I*c) - 3*I*a^2*d*e^(4*I*d*x + 4
*I*c))*sqrt(1/(a^3*d^2))*log((2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*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(1/2)*(3*I*a^2*d*e^(6*I
*d*x + 6*I*c) + 3*I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^3*d^2))*log(-(2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*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^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x + 4*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 )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

Integral(tan(c + d*x)**4/(a*(I*tan(c + d*x) + 1))**(3/2), 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}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^4/(I*a*tan(d*x + c) + a)^(3/2), x)