3.2.27 \(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [127]

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

[Out]

-1/8*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+151/60*I/a^2/d/(a+I*a*tan(d*x+c
))^(1/2)+83/30*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d-1/5*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(5/2)+17/30*I*tan(d*x+c)
^2/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.25, antiderivative size = 176, 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 = {3639, 3676, 3673, 3607, 3561, 212} \begin {gather*} -\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 {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 {\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4*I)*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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3561

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 3607

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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(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 3673

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 3676

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))^{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 \text {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*}

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Mathematica [A]
time = 1.61, size = 149, normalized size = 0.85 \begin {gather*} \frac {e^{-6 i (c+d x)} \left (-\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)}}}+i \left (1+e^{2 i (c+d x)}\right ) \left (3-26 e^{2 i (c+d x)}+194 e^{4 i (c+d x)}+463 e^{6 i (c+d x)}\right ) \sec ^2(c+d x)\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

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 = 0.17, size = 111, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}\right )}{d \,a^{3}}\)	\(111\)
default	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}\right )}{d \,a^{3}}\)	\(111\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 139, normalized size = 0.79 \begin {gather*} \frac {i \, {\left (15 \, \sqrt {2} a^{\frac {5}{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 ) + 480 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2} + \frac {4 \, {\left (255 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 70 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 12 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{5} d} \end {gather*}

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]

1/240*I*(15*sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a))) + 480*sqrt(I*a*tan(d*x + c) + a)*a^2 + 4*(255*(I*a*tan(d*x + c) + a)^2*a^3 - 70*(I*a*tan(d*x
+ c) + a)*a^4 + 12*a^5)/(I*a*tan(d*x + c) + a)^(5/2))/(a^5*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (131) = 262\).
time = 0.38, size = 283, normalized size = 1.61 \begin {gather*} \frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a 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 (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} 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 (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 )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

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/(I*a*(tan(c + d*x) - I))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.25, size = 129, normalized size = 0.73 \begin {gather*} \frac {\frac {1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{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}}{8\,{\left (-a\right )}^{5/2}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(1i/(5*d) + ((a + a*tan(c + d*x)*1i)^2*17i)/(4*a^2*d) - ((a + a*tan(c + d*x)*1i)*7i)/(6*a*d))/(a + a*tan(c + d
*x)*1i)^(5/2) + ((a + a*tan(c + d*x)*1i)^(1/2)*2i)/(a^3*d) + (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1
/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d)

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