3.127 \(\int \frac{\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 a^3 d}+\frac{151 i}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((-I/4)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) - Tan[c + d*x]^3/(5*d*(a +
I*a*Tan[c + d*x])^(5/2)) + (((17*I)/30)*Tan[c + d*x]^2)/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((151*I)/60)/(a^2
*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((83*I)/30)*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d)

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Rubi [A]  time = 0.370926, antiderivative size = 176, 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, 3526, 3480, 206} \[ \frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 a^3 d}+\frac{151 i}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-I/4)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) - Tan[c + d*x]^3/(5*d*(a +
I*a*Tan[c + d*x])^(5/2)) + (((17*I)/30)*Tan[c + d*x]^2)/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((151*I)/60)/(a^2
*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((83*I)/30)*Sqrt[a + I*a*Tan[c + d*x]])/(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 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, 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))^{5/2}} \, dx &=-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac{\int \frac{\tan ^2(c+d x) \left (-3 a+\frac{11}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\tan (c+d x) \left (-17 i a^2-\frac{83}{4} a^2 \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 a^3 d}+\frac{\int \frac{\frac{83 a^2}{4}-17 i a^2 \tan (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 i}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 a^3 d}+\frac{\int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 i}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 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 )}{4 a^2 d}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 i}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{83 i \sqrt{a+i a \tan (c+d x)}}{30 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.51176, size = 149, normalized size = 0.85 \[ \frac{e^{-6 i (c+d x)} \left (i \left (1+e^{2 i (c+d x)}\right ) \left (-26 e^{2 i (c+d x)}+194 e^{4 i (c+d x)}+463 e^{6 i (c+d x)}+3\right ) \sec ^2(c+d x)-\frac{60 i e^{7 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{240 a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-60*I)*E^((7*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + I*(1 + E^((2*I)*(c + d
*x)))*(3 - 26*E^((2*I)*(c + d*x)) + 194*E^((4*I)*(c + d*x)) + 463*E^((6*I)*(c + d*x)))*Sec[c + d*x]^2)/(240*a^
2*d*E^((6*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.024, size = 111, normalized size = 0.6 \begin{align*}{\frac{2\,i}{d{a}^{3}} \left ( \sqrt{a+ia\tan \left ( dx+c \right ) }-{\frac{\sqrt{2}}{16}\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) }-{\frac{7\,{a}^{2}}{12} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{17\,a}{8}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{{a}^{3}}{10} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{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))^(5/2),x)

[Out]

2*I/d/a^3*((a+I*a*tan(d*x+c))^(1/2)-1/16*a^(1/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))
-7/12/(a+I*a*tan(d*x+c))^(3/2)*a^2+17/8/(a+I*a*tan(d*x+c))^(1/2)*a+1/10*a^3/(a+I*a*tan(d*x+c))^(5/2))

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.23203, size = 865, normalized size = 4.91 \begin{align*} \frac{{\left (-15 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} 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 ) + 15 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{1}{a^{5} 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} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (463 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 194 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 26 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} d} \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))^(5/2),x, algorithm="fricas")

[Out]

1/120*(-15*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*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)) + 15*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt(1/(a^5*
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)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(463*I*e^(6*I*d*x + 6*I*c) + 194*I*e^(4*I*d*x
+ 4*I*c) - 26*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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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{5}{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))**(5/2),x)

[Out]

Integral(tan(c + d*x)**4/(a*(I*tan(c + d*x) + 1))**(5/2), 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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)