3.2.78 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (\frac {3 b B}{2 a}+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [178]

Optimal. Leaf size=190 \[ \frac {2 (-1)^{3/4} a^{5/2} B \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2+2 i) a^{3/2} (2 a+3 i b) B \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.48, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3674, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {2 (-1)^{3/4} a^{5/2} B \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2+2 i) a^{3/2} B (2 a+3 i b) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a B (a+3 i b) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 3674

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

Rule 3680

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

Rule 3682

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} (a+3 i b) B+\frac {3}{2} i a B \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4}{3} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {9}{4} a (i a-2 b) B-\frac {3}{4} a^2 B \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}-(i a B) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+(2 a (2 i a-3 b) B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (i a^3 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (4 a^3 (2 a+3 i b) B\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} (2 a+3 i b) B \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 (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (2 i a^3 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(2+2 i) a^{3/2} (2 a+3 i b) B \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 (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (2 i a^3 B\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {2 (-1)^{3/4} a^{5/2} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2+2 i) a^{3/2} (2 a+3 i b) B \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 (a+3 i b) B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {b B (a+i a \tan (c+d x))^{3/2}}{d \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 6.68, size = 326, normalized size = 1.72 \begin {gather*} \frac {\left (-\frac {2 i \sqrt {2} e^{-2 i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (-2 (2 a+3 i b) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+\sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {2 (2 a+7 i b+b \cot (c+d x)) \sqrt {\sec (c+d x)} (\cos (2 c)-i \sin (2 c))}{(\cos (d x)+i \sin (d x))^2 \sqrt {\tan (c+d x)}}\right ) (a+i a \tan (c+d x))^{5/2} \left (\frac {3 b B}{2 a}+B \tan (c+d x)\right )}{d \sec ^{\frac {7}{2}}(c+d x) (3 b \cos (c+d x)+2 a \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((((-2*I)*Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-2*(2*a + (3*I)*b)*ArcTan
h[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*a*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((
2*I)*(c + d*x))]]))/(E^((2*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c
+ d*x)))]) - (2*(2*a + (7*I)*b + b*Cot[c + d*x])*Sqrt[Sec[c + d*x]]*(Cos[2*c] - I*Sin[2*c]))/((Cos[d*x] + I*Si
n[d*x])^2*Sqrt[Tan[c + d*x]]))*(a + I*a*Tan[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/(d*Sec[c + d*x]^
(7/2)*(3*b*Cos[c + d*x] + 2*a*Sin[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. 619 vs. \(2 (156 ) = 312\).
time = 0.18, size = 620, normalized size = 3.26

method	result	size
derivativedivides	\(\frac {B a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (6 i \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^{2} \left (\tan ^{2}\left (d x +c \right )\right )-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 ^{2}\left (d x +c \right )\right )+2 i \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 ^{2}\left (d x +c \right )\right )-14 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, b -12 \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 b \left (\tan ^{2}\left (d x +c \right )\right )-\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 ^{2}\left (d x +c \right )\right )-2 \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 ^{2}\left (d x +c \right )\right )-4 a \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-2 b \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 )}{2 d \tan \left (d x +c \right )^{\frac {3}{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}}\)	\(620\)
default	\(\frac {B a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (6 i \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^{2} \left (\tan ^{2}\left (d x +c \right )\right )-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 ^{2}\left (d x +c \right )\right )+2 i \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 ^{2}\left (d x +c \right )\right )-14 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, b -12 \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 b \left (\tan ^{2}\left (d x +c \right )\right )-\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 ^{2}\left (d x +c \right )\right )-2 \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 ^{2}\left (d x +c \right )\right )-4 a \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-2 b \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 )}{2 d \tan \left (d x +c \right )^{\frac {3}{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}}\)	\(620\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*B*a*(a*(1+I*tan(d*x+c)))^(1/2)*(6*I*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^2*tan(d*x+c)^2-I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*t
an(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+2*I*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)^2-14*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)*b-12*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*b*tan(d*x+c)^2-(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)^2-2*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)^2-4*a*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)-2*b*(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)^(3/2)/(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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*integrate((2*B*tan(d*x + c) + 3*B*b/a)*(I*a*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(5/2), x)

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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. 900 vs. \(2 (146) = 292\).
time = 0.60, size = 900, normalized size = 4.74 \begin {gather*} \frac {2 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {-4 i \, B^{2} a^{5} + 12 \, B^{2} a^{4} b + 9 i \, B^{2} a^{3} b^{2}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} d \sqrt {-\frac {-4 i \, B^{2} a^{5} + 12 \, B^{2} a^{4} b + 9 i \, B^{2} a^{3} b^{2}}{d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left (2 i \, B a^{2} - 3 \, B a b + {\left (2 i \, B a^{2} - 3 \, B a b\right )} e^{\left (2 i \, d x + 2 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, B a^{2} - 3 \, B a b}\right ) - 2 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {-4 i \, B^{2} a^{5} + 12 \, B^{2} a^{4} b + 9 i \, B^{2} a^{3} b^{2}}{d^{2}}} \log \left (-\frac {{\left (\sqrt {2} d \sqrt {-\frac {-4 i \, B^{2} a^{5} + 12 \, B^{2} a^{4} b + 9 i \, B^{2} a^{3} b^{2}}{d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left (2 i \, B a^{2} - 3 \, B a b + {\left (2 i \, B a^{2} - 3 \, B a b\right )} e^{\left (2 i \, d x + 2 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, B a^{2} - 3 \, B a b}\right ) + 4 \, \sqrt {2} {\left (B a b e^{\left (3 i \, d x + 3 i \, c\right )} - {\left (i \, B a^{2} - 4 \, B a b\right )} e^{\left (5 i \, d x + 5 i \, c\right )} - {\left (-i \, B a^{2} + 3 \, B a b\right )} 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}} + \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (\sqrt {2} {\left (B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + B a^{2}\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 {4 i \, B^{2} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{B a^{2}}\right ) - \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (\sqrt {2} {\left (B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + B a^{2}\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 {4 i \, B^{2} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{B a^{2}}\right )}{2 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-4*I*B^2*a^5 + 12*B^2*a^4*b + 9*I*
B^2*a^3*b^2)/d^2)*log((sqrt(2)*d*sqrt(-(-4*I*B^2*a^5 + 12*B^2*a^4*b + 9*I*B^2*a^3*b^2)/d^2)*e^(I*d*x + I*c) +
sqrt(2)*(2*I*B*a^2 - 3*B*a*b + (2*I*B*a^2 - 3*B*a*b)*e^(2*I*d*x + 2*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq
rt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(2*I*B*a^2 - 3*B*a*b)) - 2*sqrt(2
)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-4*I*B^2*a^5 + 12*B^2*a^4*b + 9*I*B^2*a^3*b^2)/
d^2)*log(-(sqrt(2)*d*sqrt(-(-4*I*B^2*a^5 + 12*B^2*a^4*b + 9*I*B^2*a^3*b^2)/d^2)*e^(I*d*x + I*c) - sqrt(2)*(2*I
*B*a^2 - 3*B*a*b + (2*I*B*a^2 - 3*B*a*b)*e^(2*I*d*x + 2*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)))*e^(-I*d*x - I*c)/(2*I*B*a^2 - 3*B*a*b)) + 4*sqrt(2)*(B*a*b*e^(
3*I*d*x + 3*I*c) - (I*B*a^2 - 4*B*a*b)*e^(5*I*d*x + 5*I*c) - (-I*B*a^2 + 3*B*a*b)*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)) + sqrt(4*I*B^2*a^5/d^2)*(d
*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*c) + B*a^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)) + I*sqrt(4*I*B^2*a^5/d^2
)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(B*a^2)) - sqrt(4*I*B^2*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x
+ 2*I*c) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*c) + B*a^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)) - I*sqrt(4*I*B^2*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)
/(B*a^2)))/(d*e^(4*I*d*x + 4*I*c) - 2*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*} \frac {B \left (\int \frac {2 a^{3} \sqrt {i a \tan {\left (c + d x \right )} + a}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- 2 a^{3} \sqrt {i a \tan {\left (c + d x \right )} + a} \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int \frac {4 i a^{3} \sqrt {i a \tan {\left (c + d x \right )} + a}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \frac {3 a^{2} b \sqrt {i a \tan {\left (c + d x \right )} + a}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 a^{2} b \sqrt {i a \tan {\left (c + d x \right )} + a}}{\sqrt {\tan {\left (c + d x \right )}}}\right )\, dx + \int \frac {6 i a^{2} b \sqrt {i a \tan {\left (c + d x \right )} + a}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right )}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(Integral(2*a**3*sqrt(I*a*tan(c + d*x) + a)/tan(c + d*x)**(3/2), x) + Integral(-2*a**3*sqrt(I*a*tan(c + d*x)
+ a)*sqrt(tan(c + d*x)), x) + Integral(4*I*a**3*sqrt(I*a*tan(c + d*x) + a)/sqrt(tan(c + d*x)), x) + Integral(
3*a**2*b*sqrt(I*a*tan(c + d*x) + a)/tan(c + d*x)**(5/2), x) + Integral(-3*a**2*b*sqrt(I*a*tan(c + d*x) + a)/sq
rt(tan(c + d*x)), x) + Integral(6*I*a**2*b*sqrt(I*a*tan(c + d*x) + a)/tan(c + d*x)**(3/2), x))/(2*a)

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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))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),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 (B\,\mathrm {tan}\left (c+d\,x\right )+\frac {3\,B\,b}{2\,a}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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