\(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx\) [564]

Optimal result

Integrand size = 23, antiderivative size = 223 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3624, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[In]

Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^2,x]

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d
*x]])/(2*Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) +
(4*a*b*Sqrt[Tan[c + d*x]])/d + (2*b^2*Tan[c + d*x]^(3/2))/(3*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 3624

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.44 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=\frac {3 (-1)^{3/4} (a-i b)^2 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-3 (-1)^{3/4} (a+i b)^2 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} (6 a+b \tan (c+d x))}{3 d} \]

[In]

Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^2,x]

[Out]

(3*(-1)^(3/4)*(a - I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 3*(-1)^(3/4)*(a + I*b)^2*ArcTanh[(-1)^(3/4)*
Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c + d*x]]*(6*a + b*Tan[c + d*x]))/(3*d)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+4 a b \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(212\)
default	\(\frac {\frac {2 b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+4 a b \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}+\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(212\)
parts	\(\frac {a^{2} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {b^{2} \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {2 a b \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(299\)

[In]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/3*b^2*tan(d*x+c)^(3/2)+4*a*b*tan(d*x+c)^(1/2)-1/2*a*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^2-b^2)*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. 1005 vs. \(2 (189) = 378\).

Time = 0.27 (sec) , antiderivative size = 1005, normalized size of antiderivative = 4.51 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=\frac {3 \, d \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left ({\left ({\left (a^{2} - b^{2}\right )} d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} - 2 \, {\left (a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} d\right )} \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) - 3 \, d \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left (-{\left ({\left (a^{2} - b^{2}\right )} d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} - 2 \, {\left (a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} d\right )} \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} + d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) - 3 \, d \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left ({\left ({\left (a^{2} - b^{2}\right )} d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} + 2 \, {\left (a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} d\right )} \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) + 3 \, d \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} \log \left (-{\left ({\left (a^{2} - b^{2}\right )} d^{3} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}} + 2 \, {\left (a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} d\right )} \sqrt {\frac {4 \, a^{3} b - 4 \, a b^{3} - d^{2} \sqrt {-\frac {a^{8} - 12 \, a^{6} b^{2} + 38 \, a^{4} b^{4} - 12 \, a^{2} b^{6} + b^{8}}{d^{4}}}}{d^{2}}} + {\left (a^{8} - 4 \, a^{6} b^{2} - 10 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {\tan \left (d x + c\right )}\right ) + 4 \, {\left (b^{2} \tan \left (d x + c\right ) + 6 \, a b\right )} \sqrt {\tan \left (d x + c\right )}}{6 \, d} \]

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*d*sqrt((4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*log
(((a^2 - b^2)*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - 2*(a^5*b - 6*a^3*b^3 + a*b^5
)*d)*sqrt((4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a^8
- 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(tan(d*x + c))) - 3*d*sqrt((4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8
- 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*log(-((a^2 - b^2)*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a
^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - 2*(a^5*b - 6*a^3*b^3 + a*b^5)*d)*sqrt((4*a^3*b - 4*a*b^3 + d^2*sqrt(-(a^8 -
12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt
(tan(d*x + c))) - 3*d*sqrt((4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d
^4))/d^2)*log(((a^2 - b^2)*d^3*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + 2*(a^5*b - 6*a^
3*b^3 + a*b^5)*d)*sqrt((4*a^3*b - 4*a*b^3 - d^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))
/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*sqrt(tan(d*x + c))) + 3*d*sqrt((4*a^3*b - 4*a*b^3 - d
^2*sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2)*log(-((a^2 - b^2)*d^3*sqrt(-(a^8 - 12*a
^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + 2*(a^5*b - 6*a^3*b^3 + a*b^5)*d)*sqrt((4*a^3*b - 4*a*b^3 - d^2*
sqrt(-(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))/d^2) + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^
6 + b^8)*sqrt(tan(d*x + c))) + 4*(b^2*tan(d*x + c) + 6*a*b)*sqrt(tan(d*x + c)))/d

Sympy [F]

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

[In]

integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sqrt(tan(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.83 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=\frac {8 \, b^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 6 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 3 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 3 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 48 \, a b \sqrt {\tan \left (d x + c\right )}}{12 \, d} \]

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(8*b^2*tan(d*x + c)^(3/2) + 6*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x +
c)))) + 6*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 3*sqrt(2)*(a^2 +
2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 3*sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*
sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 48*a*b*sqrt(tan(d*x + c)))/d

Giac [F(-1)]

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

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.84 (sec) , antiderivative size = 954, normalized size of antiderivative = 4.28 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 \, dx=\frac {2\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {4\,a\,b\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\mathrm {atan}\left (\frac {a^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}+\frac {b^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}-\frac {a^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3\,b}{d^2}-\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}-\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,192{}\mathrm {i}}{\frac {16\,a^6}{d}-\frac {16\,b^6}{d}+\frac {112\,a^2\,b^4}{d}-\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}\right )\,\sqrt {-\frac {a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}+\frac {b^4\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,32{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}-\frac {a^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^4\,1{}\mathrm {i}}{4\,d^2}+\frac {b^4\,1{}\mathrm {i}}{4\,d^2}-\frac {a\,b^3}{d^2}+\frac {a^3\,b}{d^2}-\frac {a^2\,b^2\,3{}\mathrm {i}}{2\,d^2}}\,192{}\mathrm {i}}{\frac {16\,b^6}{d}-\frac {16\,a^6}{d}-\frac {112\,a^2\,b^4}{d}+\frac {112\,a^4\,b^2}{d}+\frac {a\,b^5\,32{}\mathrm {i}}{d}+\frac {a^5\,b\,32{}\mathrm {i}}{d}-\frac {a^3\,b^3\,192{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \]

[In]

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

[Out]

atan((a^4*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(
2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)
/d - (112*a^4*b^2)/d) + (b^4*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^
2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)
/d - (a^3*b^3*192i)/d - (112*a^4*b^2)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^
3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d + (a^5*
b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - (112*a^4*b^2)/d))*(-(4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2
*b^2*6i)/(4*d^2))^(1/2)*2i - atan((a^4*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 +
(a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d - (11
2*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d) + (b^4*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d
^2) - (a*b^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i)/d +
(a^5*b*32i)/d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a^4*1i)/
(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((16*b^6)/d - (16*a
^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d))*((4*a^3*b - 4*a
*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)/(4*d^2))^(1/2)*2i + (2*b^2*tan(c + d*x)^(3/2))/(3*d) + (4*a*b*tan(c + d*x
)^(1/2))/d