\(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [379]

Optimal result

Integrand size = 31, antiderivative size = 254 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[Out]

-1/2*(b*(A-B)+a*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(b*(A-B)+a*(A+B))*arctan(1+2^(1/2)*ta
n(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a*(A-B)-b*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1/4*(a*(A-
B)-b*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2*(A*a-B*b)*tan(d*x+c)^(1/2)/d+2/3*(A*b+B*a)*t
an(d*x+c)^(3/2)/d+2/5*b*B*tan(d*x+c)^(5/2)/d

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3673, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(a (A+B)+b (A-B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 (a B+A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[In]

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

[Out]

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

Rule 210

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

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3673

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\int \tan ^{\frac {3}{2}}(c+d x) (a A-b B+(A b+a B) \tan (c+d x)) \, dx \\ & = \frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\int \sqrt {\tan (c+d x)} (-A b-a B+(a A-b B) \tan (c+d x)) \, dx \\ & = \frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\int \frac {-a A+b B-(A b+a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 \text {Subst}\left (\int \frac {-a A+b B+(-A b-a B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d} \\ & = \frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d} \\ & = \frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 (a A-b B) \sqrt {\tan (c+d x)}}{d}+\frac {2 (A b+a B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.53 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {15 \sqrt [4]{-1} (a-i b) (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+15 \sqrt [4]{-1} (a+i b) (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (15 (a A-b B)+5 (A b+a B) \tan (c+d x)+3 b B \tan ^2(c+d x)\right )}{15 d} \]

[In]

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

[Out]

(15*(-1)^(1/4)*(a - I*b)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 15*(-1)^(1/4)*(a + I*b)*(A + I*B)*A
rcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(15*(a*A - b*B) + 5*(A*b + a*B)*Tan[c + d*x] + 3*
b*B*Tan[c + d*x]^2))/(15*d)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {2 B b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 B a \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a A \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B b +\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(253\)
default	\(\frac {\frac {2 B b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 B a \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 a A \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B b +\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(253\)
parts	\(\frac {\left (A b +B a \right ) \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {B b \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(323\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2252 vs. \(2 (216) = 432\).

Time = 0.38 (sec) , antiderivative size = 2252, normalized size of antiderivative = 8.87 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/30*(15*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3
*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4
)/d^4))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A
^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*
A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2
)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2
+ 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3
*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c))) - 15*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A
^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*
a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)*d^3*sqrt(-((A^4
- 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^
3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (
A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4
- 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 +
B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4
)*sqrt(tan(d*x + c))) - 15*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 +
B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A
^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*
b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^3 -
A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b
^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^
2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A
^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c))) + 15*d*sqrt(-(2*A*B*a^2 -
2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 1
0*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)
*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^
3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5
*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A
^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (
A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3
- (A^4 - B^4)*b^4)*sqrt(tan(d*x + c))) - 4*(3*B*b*tan(d*x + c)^2 + 15*A*a - 15*B*b + 5*(B*a + A*b)*tan(d*x + c
))*sqrt(tan(d*x + c)))/d

Sympy [F]

\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))*tan(c + d*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.83 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {24 \, B b \tan \left (d x + c\right )^{\frac {5}{2}} - 30 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 30 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 120 \, {\left (A a - B b\right )} \sqrt {\tan \left (d x + c\right )}}{60 \, d} \]

[In]

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

[Out]

1/60*(24*B*b*tan(d*x + c)^(5/2) - 30*sqrt(2)*((A + B)*a + (A - B)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(
d*x + c)))) - 30*sqrt(2)*((A + B)*a + (A - B)*b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 15*sq
rt(2)*((A - B)*a - (A + B)*b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 15*sqrt(2)*((A - B)*a - (A
+ B)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 40*(B*a + A*b)*tan(d*x + c)^(3/2) + 120*(A*a - B
*b)*sqrt(tan(d*x + c)))/d

Giac [F(-1)]

Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 14.41 (sec) , antiderivative size = 1492, normalized size of antiderivative = 5.87 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

atan((A^2*a^2*tan(c + d*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2
*d^2))^(1/2)*32i)/((16*A^3*b^3)/d + (16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2))/d^3 - (16*A
^3*a^2*b)/d) - (A^2*b^2*tan(c + d*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A
^2*a*b)/(2*d^2))^(1/2)*32i)/((16*A^3*b^3)/d + (16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2))/d
^3 - (16*A^3*a^2*b)/d))*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1
/2)*2i - atan((A^2*a^2*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (
A^2*a*b)/(2*d^2))^(1/2)*32i)/((16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2))/d^3 - (16*A^3*b^3
)/d + (16*A^3*a^2*b)/d) - (A^2*b^2*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)
/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2)*32i)/((16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2))/d^3 -
(16*A^3*b^3)/d + (16*A^3*a^2*b)/d))*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a
*b)/(2*d^2))^(1/2)*2i - atan((B^2*a^2*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4
- B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2)*32i)/((16*B^3*a^3)/d + (16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4
*d^4)^(1/2))/d^3 - (16*B^3*a*b^2)/d) - (B^2*b^2*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B
^4*b^4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2)*32i)/((16*B^3*a^3)/d + (16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4
- B^4*a^4*d^4)^(1/2))/d^3 - (16*B^3*a*b^2)/d))*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^
4*d^4)^(1/2)/(4*d^4))^(1/2)*2i + atan((B^2*a^2*tan(c + d*x)^(1/2)*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*
d^4)^(1/2)/(4*d^4) + (B^2*a*b)/(2*d^2))^(1/2)*32i)/((16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1
/2))/d^3 - (16*B^3*a^3)/d + (16*B^3*a*b^2)/d) - (B^2*b^2*tan(c + d*x)^(1/2)*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4
- B^4*a^4*d^4)^(1/2)/(4*d^4) + (B^2*a*b)/(2*d^2))^(1/2)*32i)/((16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a
^4*d^4)^(1/2))/d^3 - (16*B^3*a^3)/d + (16*B^3*a*b^2)/d))*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2
)/(4*d^4) + (B^2*a*b)/(2*d^2))^(1/2)*2i + (2*A*a*tan(c + d*x)^(1/2))/d + (2*A*b*tan(c + d*x)^(3/2))/(3*d) + (2
*B*a*tan(c + d*x)^(3/2))/(3*d) - (2*B*b*tan(c + d*x)^(1/2))/d + (2*B*b*tan(c + d*x)^(5/2))/(5*d)