3.2.28 \(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [128]

Optimal. Leaf size=133 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

1/8*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)+31/20/a^2/d/(a+I*a*tan(d*x+c))^(1/
2)-1/5*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(5/2)-13/30/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.16, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3639, 3671, 3607, 3561, 212} \begin {gather*} \frac {\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 {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(4*Sqrt[2]*a^(5/2)*d) - Tan[c + d*x]^2/(5*d*(a + I*a*Tan
[c + d*x])^(5/2)) - 13/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + 31/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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 3671

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(A*b - a*B))*(a*c + b*d)*((a + b*Tan[e + f*x])^m/(2*a^2*f*m)), x] + Dis
t[1/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B*d + 2*a*B*d*Tan[e + f*x], x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x) \left (-2 a+\frac {9}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \int \frac {-\frac {13 a^2}{2}+9 i a^2 \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{10 a^4}\\ &=-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\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 {\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 ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.20, size = 135, normalized size = 1.02 \begin {gather*} \frac {e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \left (\sqrt {1+e^{2 i (c+d x)}} \left (3-19 e^{2 i (c+d x)}+83 e^{4 i (c+d x)}\right )+15 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sec ^2(c+d x)}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(3 - 19*E^((2*I)*(c + d*x)) + 83*E^((4*I)*(c +
d*x))) + 15*E^((5*I)*(c + d*x))*ArcSinh[E^(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 = 0.19, size = 93, normalized size = 0.70

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {7}{8 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {5 a}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}\right )}{d \,a^{2}}\)	\(93\)
default	\(-\frac {2 \left (-\frac {7}{8 \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {5 a}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}\right )}{d \,a^{2}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/d/a^2*(-7/8/(a+I*a*tan(d*x+c))^(1/2)+5/12*a/(a+I*a*tan(d*x+c))^(3/2)-1/10*a^2/(a+I*a*tan(d*x+c))^(5/2)-1/16
*2^(1/2)/a^(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 = 0.51, size = 121, normalized size = 0.91 \begin {gather*} -\frac {15 \, \sqrt {2} a^{\frac {3}{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 ) - \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 50 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + 12 \, a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(15*sqrt(2)*a^(3/2)*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))) - 4*(105*(I*a*tan(d*x + c) + a)^2*a^2 - 50*(I*a*tan(d*x + c) + a)*a^3 + 12*a^4)/(I*a*tan(d*x
+ c) + a)^(5/2))/(a^4*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. \(2 (102) = 204\).
time = 0.37, size = 283, normalized size = 2.13 \begin {gather*} \frac {{\left (15 \, \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 \, \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 (83 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 64 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 16 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\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)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*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
*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)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1))*(83*e^(6*I*d*x + 6*I*c) + 64*e^(4*I*d*x + 4*I*c) - 16*e^(2*I*d*x + 2*I*c) + 3)
)*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 ^{3}{\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)**3/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(tan(c + d*x)**3/(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)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^3/(I*a*tan(d*x + c) + a)^(5/2), x)

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Mupad [B]
time = 3.86, size = 93, normalized size = 0.70 \begin {gather*} \frac {\frac {7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {5\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}+\frac {a^2}{5}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

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

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