3.2.72 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [172]

Optimal. Leaf size=196 \[ \frac {(-1)^{3/4} a^{5/2} (2 A-5 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.45, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3674, 3675, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {(-1)^{3/4} a^{5/2} (2 A-5 i 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 {(4+4 i) a^{5/2} (A-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 {a^2 (-B+2 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && 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} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (4 i A+B)+\frac {1}{2} a (2 A+i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (2 i A+B)-\frac {1}{4} a^2 (2 A-5 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\frac {1}{2} (a (2 i A+5 B)) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\left (8 a^4 (A-i 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 {\left (a^3 (2 i A+5 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {(4+4 i) a^{5/2} (A-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 {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left (a^3 (2 i A+5 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} (A-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 {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left (a^3 (2 i A+5 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 {\sqrt [4]{-1} a^{5/2} (2 i A+5 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 {(4+4 i) a^{5/2} (A-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 {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 6.89, size = 389, normalized size = 1.98 \begin {gather*} \frac {\left (\frac {\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 (32 (i A+B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )-i \sqrt {2} (2 A-5 i B) \left (\log \left (1-3 e^{2 i (c+d x)}-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )-\log \left (1-3 e^{2 i (c+d x)}+2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )\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 {8 (B+2 A \cot (c+d x)) \sqrt {\sec (c+d x)} (\cos (2 c)-i \sin (2 c)) \sqrt {\tan (c+d x)}}{(\cos (d x)+i \sin (d x))^2}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{8 d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(32*(I*A + B)*Log[E^(I*(c + d*x))
+ Sqrt[-1 + E^((2*I)*(c + d*x))]] - I*Sqrt[2]*(2*A - (5*I)*B)*(Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[2]*E^(I
*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - Log[1 - 3*E^((2*I)*(c + d*x)) + 2*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)))]) - (8*(B + 2*A*Cot[c + d*x])*Sqrt[Sec[c + d*x]]*(Cos[2*c] - I*Sin[2*c])*Sqrt[Tan[c + d*x]])
/(Cos[d*x] + I*Sin[d*x])^2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(8*d*Sec[c + d*x]^(7/2)*(A*Cos[
c + d*x] + B*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. 564 vs. \(2 (161 ) = 322\).
time = 0.12, size = 565, normalized size = 2.88

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (6 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+3 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(565\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (6 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+3 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(565\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(6*I*A*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*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)+3*B*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*B*(I*a)^(1/2
)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+4*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*(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)-4*A*(
I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-4*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))/tan(d*x+c)^(1/2)/(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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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. 857 vs. \(2 (150) = 300\).
time = 0.70, size = 857, normalized size = 4.37 \begin {gather*} -\frac {4 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 4 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 2 \, \sqrt {2} {\left ({\left (2 i \, A + B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 i \, A - B\right )} a^{2} 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 {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 5 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 5 \, B\right )} 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}} + 2 \, \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 5 \, B\right )} a^{2}}\right ) - \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 5 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 5 \, B\right )} 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}} - 2 \, \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 5 \, B\right )} a^{2}}\right )}{2 \, {\left (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)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(4*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*sqrt(-(-I*A^2
- 2*A*B + I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)*s
qrt(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
)/((-I*A - B)*a^2)) - 4*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-(sqrt
(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) - sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) +
(-I*A - 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))
)*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 2*sqrt(2)*((2*I*A + B)*a^2*e^(3*I*d*x + 3*I*c) + (2*I*A - B)*a^2*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)) + sq
rt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 5*B)*a^2*e^(2*I*d
*x + 2*I*c) + (2*I*A + 5*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)) + 2*sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A
+ 5*B)*a^2)) - sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A +
5*B)*a^2*e^(2*I*d*x + 2*I*c) + (2*I*A + 5*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)) - 2*sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*
x - I*c)/((2*I*A + 5*B)*a^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*} \int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {3}{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))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(3/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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