3.2.71 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [171]

Optimal. Leaf size=206 \[ -\frac {(-1)^{3/4} a^{5/2} (20 i A+23 B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}+\frac {(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \]

[Out]

-1/4*(-1)^(3/4)*a^(5/2)*(20*I*A+23*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-1/4*a^2*(4*A-7*I*B)*
tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/2*I*a*B*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.46, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3675, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {(-1)^{3/4} a^{5/2} (23 B+20 i A) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 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 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((-1)^(3/4)*a^(5/2)*((20*I)*A + 23*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*(4*A - (7*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(4*d) + ((I/2)*a*B*Sqrt[Tan[c +
d*x]]*(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 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))}{\sqrt {\tan (c+d x)}} \, dx &=\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{2} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (4 A-i B)+\frac {1}{2} a (4 i A+7 B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {a^2 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{2} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (4 A-3 i B)+\frac {1}{4} a^2 (20 i A+23 B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {a^2 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\left (4 a^2 (A-i B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\frac {1}{8} (a (20 A-23 i 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 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (a^3 (20 A-23 i B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (8 a^4 (i A+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 {(4+4 i) a^{5/2} (i A+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 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (a^3 (20 A-23 i B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=-\frac {(4+4 i) a^{5/2} (i A+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 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (a^3 (20 A-23 i 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 )}{4 d}\\ &=\frac {\sqrt [4]{-1} a^{5/2} (20 A-23 i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}-\frac {(4+4 i) a^{5/2} (i A+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 (4 A-7 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}\\ \end {align*}

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Mathematica [A]
time = 6.53, size = 394, normalized size = 1.91 \begin {gather*} \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \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 (128 (A-i B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )-\sqrt {2} (20 A-23 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 \sqrt {\sec (c+d x)} (\cos (2 c)-i \sin (2 c)) \sqrt {\tan (c+d x)} (4 A-9 i B+2 B \tan (c+d x))}{(\cos (d x)+i \sin (d x))^2}\right )}{32 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]))/Sqrt[Tan[c + d*x]],x]

[Out]

((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x])*((Sqrt[2]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2
*I)*(c + d*x)))]*(128*(A - I*B)*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] - Sqrt[2]*(20*A - (23*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*Sqrt[Sec[c + d*x]]*(Cos[2*c] - I*S
in[2*c])*Sqrt[Tan[c + d*x]]*(4*A - (9*I)*B + 2*B*Tan[c + d*x]))/(Cos[d*x] + I*Sin[d*x])^2))/(32*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. 565 vs. \(2 (164 ) = 328\).
time = 0.12, size = 566, normalized size = 2.75

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) a^{2} \left (-9 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +18 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+8 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -8 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+16 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -8 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +16 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{8 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(566\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) a^{2} \left (-9 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +18 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+8 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -8 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+16 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -8 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +16 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{8 d \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(566\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^(1/2)*a^2*(-9*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a+18*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)-4*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)+8*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+12*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-8*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+16*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))*a*(-I*a)^(1/2)-8*(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+16*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((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),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. 849 vs. \(2 (152) = 304\).
time = 0.76, size = 849, normalized size = 4.12 \begin {gather*} -\frac {16 \, \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 (i \, \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 ) - 16 \, \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 (-i \, \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 (4 \, A - 11 i \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (4 \, A - 7 i \, 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 (-400 i \, A^{2} - 920 \, A B + 529 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 (20 i \, A + 23 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (20 i \, A + 23 \, 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 i \, \sqrt {\frac {{\left (-400 i \, A^{2} - 920 \, A B + 529 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 (20 i \, A + 23 \, B\right )} a^{2}}\right ) - \sqrt {\frac {{\left (-400 i \, A^{2} - 920 \, A B + 529 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 (20 i \, A + 23 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (20 i \, A + 23 \, 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 i \, \sqrt {\frac {{\left (-400 i \, A^{2} - 920 \, A B + 529 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 (20 i \, A + 23 \, B\right )} a^{2}}\right )}{8 \, {\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)^(1/2),x, algorithm="fricas")

[Out]

-1/8*(16*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((I*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)) - 16*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((-I*s
qrt(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)*((4*A - 11*I*B)*a^2*e^(3*I*d*x + 3*I*c) + (4*A - 7*I*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
)) + sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((20*I*A + 23*B
)*a^2*e^(2*I*d*x + 2*I*c) + (20*I*A + 23*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*I*sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^
(-I*d*x - I*c)/((20*I*A + 23*B)*a^2)) - sqrt((-400*I*A^2 - 920*A*B + 529*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c
) + d)*log((sqrt(2)*((20*I*A + 23*B)*a^2*e^(2*I*d*x + 2*I*c) + (20*I*A + 23*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*I*sqrt((-400*I*A^2 - 920*A*B + 529*I*
B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((20*I*A + 23*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 )}{\sqrt {\tan {\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)**(1/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(5/2)*(A + B*tan(c + d*x))/sqrt(tan(c + d*x)), 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)^(1/2),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 = -0.00 \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}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,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)^(1/2),x)

[Out]

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

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