3.3.30 \(\int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [230]

Optimal. Leaf size=162 \[ \frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

(1/8-1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(5/2)/d+67/60*tan(d*x+c)^(1/2)/
a^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/5*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(5/2)+13/30*tan(d*x+c)^(1/2)/a/d/(a+I
*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.29, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3640, 3677, 12, 3625, 211} \begin {gather*} \frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 - I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + Sqrt[Tan[
c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (13*Sqrt[Tan[c + d*x]])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) +
(67*Sqrt[Tan[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*a*(b/f), Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*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]

Rule 3640

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\frac {9 a}{2}-2 i a \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {41 a^2}{4}-\frac {13}{2} i a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {15 a^3 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {13 \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {67 \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.27, size = 171, normalized size = 1.06 \begin {gather*} -\frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-3-16 e^{2 i (c+d x)}-64 e^{4 i (c+d x)}+83 e^{6 i (c+d x)}+15 e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{60 \sqrt {2} a^3 d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((-1/60*I)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-3 - 16*E^((2*I)*(c + d*x)) - 64*E^((4*I)*
(c + d*x)) + 83*E^((6*I)*(c + d*x)) + 15*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c +
d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x))*Sqrt[Tan[c + d*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 575 vs. \(2 (128 ) = 256\).
time = 0.19, size = 576, normalized size = 3.56

method	result	size
derivativedivides	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-90 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+268 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -1060 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+908 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-420 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(576\)
default	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )-90 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+268 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -1060 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+908 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-420 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(576\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^4-90*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+268*I*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3+15*I*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-1060*I*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-60*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)+908*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*tan(d*x+c)^2-420*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a^3/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(
1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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. 330 vs. \(2 (120) = 240\).
time = 0.61, size = 330, normalized size = 2.04 \begin {gather*} \frac {{\left (30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (83 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 102 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 22 \, 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(1/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(I*d*x + I*c) +
1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^
(2*I*d*x + 2*I*c) + 1)) - 30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-a^3*d*sqrt(-1/8*I/(a^5*d^2)
)*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x +
2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(83*e^(6*I*d*x + 6*I*c) + 102*e^(4*I*d*x + 4*I*c) + 22*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 {1}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {\tan {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(1/((I*a*(tan(c + d*x) - I))**(5/2)*sqrt(tan(c + d*x))), x)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1348 vs. \(2 (120) = 240\).
time = 4.83, size = 1348, normalized size = 8.32 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(I*a*sqrt(abs(a)) - abs(a)^(3/2))*log(-1/2*(I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a
)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*
tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2 + 8*sqrt(2) - 12)/(-1/2*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) +
a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan
(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2 + 4*sqrt(2) + 6))/(a^4*d) - 1/15*
sqrt(2)*(15*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c)
+ a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*ab
s(a)/a^2)^8*a*sqrt(abs(a)) + 15*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqr
t(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)
*a + a^2)/a^2) + 1)*abs(a)/a^2)^8*abs(a)^(3/2) - 480*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*a
bs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(
I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^6*a*sqrt(abs(a)) + 480*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a
)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(
d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^6*abs(a)^(3/2) - 8800*(sqrt(2)*sqrt(
I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x
+ c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^4*a*sqrt(abs(a))
- 8800*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) +
a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(
a)/a^2)^4*abs(a)^(3/2) + 20480*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt
(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*
a + a^2)/a^2) + 1)*abs(a)/a^2)^2*a*sqrt(abs(a)) - 20480*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*
abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*
(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2*abs(a)^(3/2) + 17152*a*sqrt(abs(a)) + 17152*I*abs(a)^(
3/2))/(((sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a
)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)
/a^2)^2 - 4*I)^5*a^4*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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