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

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

[Out]

-1/4*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)-10*I*(a+I*a*tan(d*x+c))^(1/2)/a
^2/d+5/2*I*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(1/2)-1/3*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+7/2*I*(a+I*a*
tan(d*x+c))^(3/2)/a^3/d

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Rubi [A]  time = 0.35, antiderivative size = 174, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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]

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*}

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Mathematica [A]  time = 1.27, 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]

((-1/24*I)*(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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fricas [B]  time = 0.44, size = 333, normalized size = 1.91 \[ \frac {\sqrt {\frac {1}{2}} {\left (-3 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - 3 i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {\frac {1}{2}} {\left (3 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-52 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 87 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

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(1/2)*(-3*I*a^2*d*e^(5*I*d*x + 5*I*c) - 3*I*a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt
(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I
*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(1/2)*(3*I*a^2*d*e^(5*I*d*x + 5*I*c) + 3*I*a^2*d*e^(3*I*d*x + 3*I*c))*sqr
t(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
*sqrt(1/(a^3*d^2)) - a*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))*(-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))/(a^2*d*e^(5*I*d*x + 5*I*c) + a^
2*d*e^(3*I*d*x + 3*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.16, size = 113, normalized size = 0.65 \[ \frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}-\frac {7 a^{2}}{4 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a^{3}}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}\right )}{d \,a^{3}} \]

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)-7/4*a^2/(a+I*a*tan(d*x+c))^(1/2)+1/6*a^3/
(a+I*a*tan(d*x+c))^(3/2)-1/8*a^(3/2)*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 [A]  time = 0.54, size = 139, normalized size = 0.80 \[ \frac {i \, {\left (3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 16 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 96 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3} - \frac {4 \, {\left (21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a^{5} d} \]

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]

1/24*I*(3*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 16*(I*a*tan(d*x + c) + a)^(3/2)*a^2 - 96*sqrt(I*a*tan(d*x + c) + a)*a^3 - 4*(21*(I*a*tan(d*x
+ c) + a)*a^4 - 2*a^5)/(I*a*tan(d*x + c) + a)^(3/2))/(a^5*d)

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mupad [B]  time = 0.28, size = 129, normalized size = 0.74 \[ \frac {\frac {1{}\mathrm {i}}{3\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

(1i/(3*d) - ((a + a*tan(c + d*x)*1i)*7i)/(2*a*d))/(a + a*tan(c + d*x)*1i)^(3/2) - ((a + a*tan(c + d*x)*1i)^(1/
2)*4i)/(a^2*d) + ((a + a*tan(c + d*x)*1i)^(3/2)*2i)/(3*a^3*d) - (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)
^(1/2))/(2*(-a)^(1/2)))*1i)/(4*(-a)^(3/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

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/(I*a*(tan(c + d*x) - I))**(3/2), x)

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