3.2.70 \(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [170]

Optimal. Leaf size=252 \[ -\frac {(-1)^{3/4} a^{5/2} (46 A-45 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 d}-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

-1/8*(-1)^(3/4)*a^(5/2)*(46*A-45*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+1/8*a^2*(18*I*A+19*B
)*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d-1/4*a^2*(2*A-3*I*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/d+
1/3*I*a*B*tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.60, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3675, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {(-1)^{3/4} a^{5/2} (46 A-45 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 )}{8 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 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {a^2 (19 B+18 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((-1)^(3/4)*a^(5/2)*(46*A - (45*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*((18*I)*A + 19*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) - (a^2*(2*A - (3*I)*B)*T
an[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/3)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/
2))/d

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 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 3678

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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && GtQ[n, 0]

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 \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{3} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (2 A-i B)+\frac {3}{2} a (2 i A+3 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{6} \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (14 A-13 i B)+\frac {3}{4} a^2 (18 i A+19 B) \tan (c+d x)\right ) \, dx\\ &=\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (18 i A+19 B)+\frac {3}{8} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 a}\\ &=\frac {a^2 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\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}{16} (a (46 i A+45 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 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\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 (46 i A+45 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 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 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (a^3 (46 i A+45 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 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 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (a^3 (46 i A+45 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 )}{8 d}\\ &=-\frac {\sqrt [4]{-1} a^{5/2} (46 i A+45 B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{8 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 (18 i A+19 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a^2 (2 A-3 i B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 6.51, size = 426, normalized size = 1.69 \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 (-256 i (A-i B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )+\sqrt {2} (46 i A+45 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 {4 \sec ^{\frac {5}{2}}(c+d x) (\cos (2 c)-i \sin (2 c)) (54 i A+49 B+(54 i A+65 B) \cos (2 (c+d x))+(-12 A+26 i B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}}{3 \cos (2 d x)+3 i \sin (2 d x)}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{64 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[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*((-256*I)*(A - I*B)*Log[E^(I*(c +
d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*((46*I)*A + 45*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)))]) + (4*Sec[c + d*x]^(5/2)*(Cos[2*c] - I*Sin[2*c])*((54*I)*A + 49*B + ((54*I)*A + 65*B)
*Cos[2*(c + d*x)] + (-12*A + (26*I)*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/(3*Cos[2*d*x] + (3*I)*Sin[2*d*x])
)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(64*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. 652 vs. \(2 (202 ) = 404\).
time = 0.12, size = 653, normalized size = 2.59

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 )}\, a^{2} \left (16 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}\, \left (\tan ^{2}\left (d x +c \right )\right )-52 i 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}\, \tan \left (d x +c \right )+54 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 -108 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+24 A \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 )-48 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 +57 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 -114 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}+96 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 -48 \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 -96 \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 \right )}{48 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(653\)
default	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (16 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}\, \left (\tan ^{2}\left (d x +c \right )\right )-52 i 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}\, \tan \left (d x +c \right )+54 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 -108 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+24 A \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 )-48 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 +57 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 -114 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}+96 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 -48 \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 -96 \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 \right )}{48 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(653\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(16*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)^2-52*I*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)+5
4*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-108*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+24*A*(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)-48*I*(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+57*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-114*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
*(I*a)^(1/2)*(-I*a)^(1/2)+96*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-48*(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-96*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)/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*sqrt(tan(d*x + c)), 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. 932 vs. \(2 (188) = 376\).
time = 0.79, size = 932, normalized size = 3.70 \begin {gather*} \frac {96 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} 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} \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 ) - 96 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} 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} \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 (66 i \, A + 91 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, {\left (-54 i \, A - 49 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, {\left (-14 i \, A - 13 \, 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}} + 3 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} 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 ({\left (46 i \, A + 45 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (46 i \, A + 45 \, 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 (2116 i \, A^{2} + 4140 \, A B - 2025 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 (46 i \, A + 45 \, B\right )} a^{2}}\right ) - 3 \, \sqrt {\frac {{\left (2116 i \, A^{2} + 4140 \, A B - 2025 i \, B^{2}\right )} 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 ({\left (46 i \, A + 45 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (46 i \, A + 45 \, 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 (2116 i \, A^{2} + 4140 \, A B - 2025 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 (46 i \, A + 45 \, B\right )} a^{2}}\right )}{48 \, {\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(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(96*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + 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)*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)) - 96*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + 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)*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)*((66*I*
A + 91*B)*a^2*e^(5*I*d*x + 5*I*c) - 2*(-54*I*A - 49*B)*a^2*e^(3*I*d*x + 3*I*c) - 3*(-14*I*A - 13*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)) + 3*
sqrt((2116*I*A^2 + 4140*A*B - 2025*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)*((46*I*A + 45*B)*a^2*e^(2*I*d*x + 2*I*c) + (46*I*A + 45*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((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2
)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((46*I*A + 45*B)*a^2)) - 3*sqrt((2116*I*A^2 + 4140*A*B - 2025*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)*((46*I*A + 45*B)*a^2*e^(2*I*d*x + 2*I
*c) + (46*I*A + 45*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((2116*I*A^2 + 4140*A*B - 2025*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((46*I*
A + 45*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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3877 deep

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 1.8Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{poly1[-16*i,0]:[1,0,-2]%%
},[0]%%%},0

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\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} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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