3.2.99 \(\int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [199]

Optimal. Leaf size=198 \[ -\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}} \]

[Out]

(-2-2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+12/5*a*(a+I*a*tan(d*x+c))^
(1/2)/d/tan(d*x+c)^(1/2)-2/5*a^2/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2)-2/5*I*a^2/d/(a+I*a*tan(d*x+c))^(1
/2)/tan(d*x+c)^(3/2)-4/5*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

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Rubi [A]
time = 0.36, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3634, 3677, 3679, 12, 3625, 211} \begin {gather*} -\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]

[Out]

((-2 - 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^2)/(5*d
*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*a^2)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d
*x]]) - (((4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) + (12*a*Sqrt[a + I*a*Tan[c + d*x]])/(5
*d*Sqrt[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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(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] && GtQ[m, 1] && Lt
Q[n, -1] && (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]

Rule 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2}{5} \int \frac {-\frac {9 i a^2}{2}+\frac {11}{2} a^2 \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-3 i a^3+2 a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{5 a^2}\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {9 a^4}{2}+3 i a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^3}\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {15 i a^5 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}}-(2 i a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}}-\frac {\left (4 a^3\right ) \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 )}{d}\\ &=-\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2}{5 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{5 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {12 a \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.79, size = 166, normalized size = 0.84 \begin {gather*} \frac {2 i a e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (e^{i (c+d x)} \left (5-10 e^{2 i (c+d x)}+9 e^{4 i (c+d x)}\right )-5 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d \left (-1+e^{2 i (c+d x)}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]

[Out]

(((2*I)/5)*a*(1 + E^((2*I)*(c + d*x)))*(E^(I*(c + d*x))*(5 - 10*E^((2*I)*(c + d*x)) + 9*E^((4*I)*(c + d*x))) -
5*(-1 + E^((2*I)*(c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Tan[c + d*x]
]*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^3)

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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. 411 vs. \(2 (159 ) = 318\).
time = 0.17, size = 412, normalized size = 2.08

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (5 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+5 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+20 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{3}\left (d x +c \right )\right )-8 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+24 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{10 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(412\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (5 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+5 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )+20 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{3}\left (d x +c \right )\right )-8 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+24 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{10 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(412\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/10/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(5*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3+5*(I*a)^(1/2)*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+20*ln(1/2*(
2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)
^3-8*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+24*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(
1/2))/tan(d*x+c)^(5/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1198 vs. \(2 (148) = 296\).
time = 0.78, size = 1198, normalized size = 6.05 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/15*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((15*I - 15)*a*cos(3*d*x + 3*c
) - (16*I - 16)*a*cos(d*x + c) - (15*I + 15)*a*sin(3*d*x + 3*c) + (16*I + 16)*a*sin(d*x + c))*cos(3/2*arctan2(
sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(15*I + 15)*a*cos(3*d*x + 3*c) + (16*I + 16)*a*cos(d*x + c) - (1
5*I - 15)*a*sin(3*d*x + 3*c) + (16*I - 16)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c)
+ 1)))*sqrt(a) + 15*(2*(-(I - 1)*a*cos(2*d*x + 2*c)^2 - (I - 1)*a*sin(2*d*x + 2*c)^2 + (2*I - 2)*a*cos(2*d*x
+ 2*c) - (I - 1)*a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*
cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + ((I +
1)*a*cos(2*d*x + 2*c)^2 + (I + 1)*a*sin(2*d*x + 2*c)^2 - (2*I + 2)*a*cos(2*d*x + 2*c) + (I + 1)*a)*log(cos(d*x
+ c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arc
tan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2
) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*
c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) +
1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 2*((-(15*I - 15)*a*c
os(5*d*x + 5*c) + (5*I - 5)*a*cos(3*d*x + 3*c) - (2*I - 2)*a*cos(d*x + c) + (15*I + 15)*a*sin(5*d*x + 5*c) - (
5*I + 5)*a*sin(3*d*x + 3*c) + (2*I + 2)*a*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) +
1)) + 3*(((I - 1)*a*cos(d*x + c) - (I + 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((I - 1)*a*cos(d*x + c) - (I +
1)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*(-(I - 1)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)
+ (I - 1)*a*cos(d*x + c) - (I + 1)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))
+ ((15*I + 15)*a*cos(5*d*x + 5*c) - (5*I + 5)*a*cos(3*d*x + 3*c) + (2*I + 2)*a*cos(d*x + c) + (15*I - 15)*a*si
n(5*d*x + 5*c) - (5*I - 5)*a*sin(3*d*x + 3*c) + (2*I - 2)*a*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), -c
os(2*d*x + 2*c) + 1)) + 3*((-(I + 1)*a*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(I + 1)*a
*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*((I + 1)*a*cos(d*x + c) + (I - 1)*a*sin(d*x + c
))*cos(2*d*x + 2*c) - (I + 1)*a*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(
2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*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. 450 vs. \(2 (148) = 296\).
time = 0.39, size = 450, normalized size = 2.27 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (-9 i \, a e^{\left (7 i \, d x + 7 i \, c\right )} + i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 5 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 5 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \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}} - 5 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 5 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{10 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/10*(4*sqrt(2)*(-9*I*a*e^(7*I*d*x + 7*I*c) + I*a*e^(5*I*d*x + 5*I*c) + 5*I*a*e^(3*I*d*x + 3*I*c) - 5*I*a*e^(
I*d*x + I*c))*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)) -
5*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(
2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) + I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + 5*(d*e^(6*I*d*x + 6*I*c)
- 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x +
2*I*c) + a)*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)) -
I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) +
3*d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(7/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)/tan(c + d*x)**(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(7/2), x)

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