\(\int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [318]

Optimal result

Integrand size = 33, antiderivative size = 186 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {\sqrt {a+i b} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d} \]

[Out]

(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d-(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(
a+I*b)^(1/2))*(a+I*b)^(1/2)/d-2*B*(a+b*tan(d*x+c))^(1/2)/d+2/15*(5*A*b-2*B*a)*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/5
*B*tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)/b/d

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3688, 3711, 3609, 3620, 3618, 65, 214} \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {\sqrt {a+i b} (-B+i A) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{d} \]

[In]

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

[Out]

(Sqrt[a - I*b]*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d - (Sqrt[a + I*b]*(I*A - B)*ArcTanh
[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d - (2*B*Sqrt[a + b*Tan[c + d*x]])/d + (2*(5*A*b - 2*a*B)*(a + b*Tan
[c + d*x])^(3/2))/(15*b^2*d) + (2*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*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 214

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

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 3618

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

Rule 3620

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

Rule 3688

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 - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {2 \int \sqrt {a+b \tan (c+d x)} \left (-a B-\frac {5}{2} b B \tan (c+d x)+\frac {1}{2} (5 A b-2 a B) \tan ^2(c+d x)\right ) \, dx}{5 b} \\ & = \frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {2 \int \sqrt {a+b \tan (c+d x)} \left (-\frac {5 A b}{2}-\frac {5}{2} b B \tan (c+d x)\right ) \, dx}{5 b} \\ & = -\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {2 \int \frac {-\frac {5}{2} b (a A-b B)-\frac {5}{2} b (A b+a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{5 b} \\ & = -\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {1}{2} ((a-i b) (A-i B)) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} ((a+i b) (A+i B)) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {(i (a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac {((i a-b) (A+i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = -\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{-1-\frac {i a}{b}+\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d}+\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{-1+\frac {i a}{b}-\frac {i x^2}{b}} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{b d} \\ & = \frac {\sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {\sqrt {a+i b} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.91 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {15 \sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+15 \sqrt {a+i b} (-i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (5 a A b-2 a^2 B-15 b^2 B+b (5 A b+a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{b^2}}{15 d} \]

[In]

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

[Out]

(15*Sqrt[a - I*b]*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + 15*Sqrt[a + I*b]*((-I)*A + B)*Ar
cTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + (2*Sqrt[a + b*Tan[c + d*x]]*(5*a*A*b - 2*a^2*B - 15*b^2*B + b*
(5*A*b + a*B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/b^2)/(15*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(872\) vs. \(2(158)=316\).

Time = 0.13 (sec) , antiderivative size = 873, normalized size of antiderivative = 4.69

method	result	size
parts	\(A \left (\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )+\frac {2 B \left (\frac {\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\sqrt {a +b \tan \left (d x +c \right )}\, b^{2}-b^{2} \left (-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )\right )}{d \,b^{2}}\)	\(873\)
derivativedivides	\(\text {Expression too large to display}\)	\(1032\)
default	\(\text {Expression too large to display}\)	\(1032\)

[In]

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

[Out]

A*(2/3/d/b*(a+b*tan(d*x+c))^(3/2)-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(
1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*ln(b
*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))-1/d*b/(2*(a^2+b^2)^(1/2)-2
*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d
/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2
+b^2)^(1/2))-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2
)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*
a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)))+2*B/d/b^2*(1/5*(a+b*tan(d*x+c))^(5/2)-1/3*a
*(a+b*tan(d*x+c))^(3/2)-(a+b*tan(d*x+c))^(1/2)*b^2-b^2*(-1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(
a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/2*(a-(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2
)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/
8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^
2)^(1/2))+1/2*((a^2+b^2)^(1/2)-a)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*t
an(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1265 vs. \(2 (152) = 304\).

Time = 0.27 (sec) , antiderivative size = 1265, normalized size of antiderivative = 6.80 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^2, x)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 25.43 (sec) , antiderivative size = 938, normalized size of antiderivative = 5.04 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}-\left (\frac {2\,B\,\left (a^2+b^2\right )}{b^2\,d}-\frac {2\,B\,a^2}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}-\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}+\frac {2\,A\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b\,d}+\frac {2\,B\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\frac {2\,B\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}-\mathrm {atan}\left (\frac {B^2\,b^4\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}+\frac {a\,b^2\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {B^2\,b^4\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}-\frac {a\,b^2\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i} \]

[In]

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

[Out]

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