3.4.0 ∫ tan3 (c+dx)(A+B tan(c+dx)) ∘ a+b tan(c+dx) dx [343]

Integrand size = 33, antiderivative size = 213 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

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

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

*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/5*B*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^2/b/d

Rubi [A] (veriﬁed)

Time = 0.64 (sec) , antiderivative size = 213, 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, 3728, 3711, 3620, 3618, 65, 214} \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

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

((A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ((A + I*B)*ArcTanh[Sqrt[a + b*

Tan[c + d*x]]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) - (2*(10*a*A*b - 8*a^2*B + 15*b^2*B)*Sqrt[a + b*Tan[c + d*x]])

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

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,

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

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]

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]

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])))

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]

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*

tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d

*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +

d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f

*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}

, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !Intege

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {2 \int \frac {\tan (c+d x) \left (-2 a B-\frac {5}{2} b B \tan (c+d x)+\frac {1}{2} (5 A b-4 a B) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{5 b} \\ & = \frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {4 \int \frac {-\frac {1}{2} a (5 A b-4 a B)-\frac {15}{4} A b^2 \tan (c+d x)-\frac {1}{4} \left (10 a A b-8 a^2 B+15 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{15 b^2} \\ & = -\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {4 \int \frac {\frac {15 b^2 B}{4}-\frac {15}{4} A b^2 \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{15 b^2} \\ & = -\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {1}{2} (-i A+B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} (i A+B) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = -\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {(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 {(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 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {(i A-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 {(i A+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) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 4.34 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {15 (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {15 (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-10 a A b+8 a^2 B-15 b^2 B+b (5 A b-4 a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{b^3}}{15 d} \]

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

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

b*Tan[c + d*x]]/Sqrt[a + I*b]])/Sqrt[a + I*b] + (2*Sqrt[a + b*Tan[c + d*x]]*(-10*a*A*b + 8*a^2*B - 15*b^2*B +

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

Maple [B] (veriﬁed)

Leaf count of result is larger than twice the leaf count of optimal. \(2014\) vs. \(2(183)=366\).

Time = 0.18 (sec) , antiderivative size = 2015, normalized size of antiderivative = 9.46


method	result	size

parts	\(\text {Expression too large to display}\)	\(2015\)
derivativedivides	\(\text {Expression too large to display}\)	\(4107\)
default	\(\text {Expression too large to display}\)	\(4107\)

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

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

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

n(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+2*((a^2+b^2)^(1/2)-a)/(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)))))+B*(2/5/

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

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

1/4/d*b/(a^2+b^2)^(3/2)*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)

)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/d/b/(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))*a^2-1/d*b/(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/d/b/(a^2+b^2)^(3/2)/(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))*a^4+3/d*b/(a^2+b^2)^(3/2)/(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))*a^2+2/d*b^3/(a^2+b^2)^(3/2)/(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))-1/4/d/b/(a^2+b^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*b/(a^2+b^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/4/d/b/(a^2+b^2)^(3/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/4/d

*b/(a^2+b^2)^(3/2)*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))*(2*

(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/d/b/(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))*a^2-1/d*b/(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/d

/b/(a^2+b^2)^(3/2)/(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))*a^4+3/d*b/(a^2+b^2)^(3/2)/(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))*a^2+2/d*b^3/(a^2+b^2)^(3/2)/(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)

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1767 vs. \(2 (177) = 354\).

Time = 0.30 (sec) , antiderivative size = 1767, normalized size of antiderivative = 8.30 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

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

^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) + ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)

*a*b - (A^2*B - B^3)*b^2)*d)*sqrt(((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt

(((a^2 + b^2)*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)/((a^4 + 2*a^2*b^

2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4)*b)*sqrt(b*

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

*b^2)*d)*sqrt(((a^2 + b^2)*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)/((a

^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt(-((a^2 + b^2)*d^2*s

qrt(-(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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - 2*

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

((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2*B - B^3)*b^2)*d)*sqrt(-((a^

2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b

^4)*d^4)) - 2*A*B*b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) + 15*b^3*d*sqrt(-((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - 2*A*B*b - (A^2 - B^2

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

+ B*a*b^2 - A*b^3)*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)/((a^4 + 2*

a^2*b^2 + b^4)*d^4)) + (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2*B - B^3)*b^2)*d)*sqrt(-((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - 2*A*B*

b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - 4*(3*B*b^2*tan(d*x + c)^2 + 8*B*a^2 - 10*A*a*b - 15*B*b^2 - (4*B*a*b

- 5*A*b^2)*tan(d*x + c))*sqrt(b*tan(d*x + c) + a))/(b^3*d)

Sympy [F]

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

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(1/2),x)

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

Maxima [F]

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

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

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

Giac [F(-1)]

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

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")

Mupad [B] (veriﬁcation not implemented)

Time = 16.40 (sec) , antiderivative size = 3054, normalized size of antiderivative = 14.34 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

int((tan(c + d*x)^3*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(1/2),x)

atan((B^2*b^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(

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

/(a^2*d^5 + b^2*d^5)) + (a*b^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b

^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-16*B^4*b^2*d^4)^(1/2)*8i)/((16*B^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4)

- (4*B*b^5*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3*b^3*d^5)/(a^2*d^4 + b^2*d^4) - (4*B

*a^2*b^3*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (B^2*a^2*b^2*d^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^

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

d^5)/(a^2*d^4 + b^2*d^4) - (4*B*b^5*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3*b^3*d^5)/(a

^2*d^4 + b^2*d^4) - (4*B*a^2*b^3*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)))*((-16*B^4*b^2*d^4)^(1/2)/(

16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*2i - atan((A^2*b^2*((A^2*a*d^2)/(4*(a^2*d

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

A^3*b^2)/d - (16*A^3*a^2*b^2*d^3)/(a^2*d^4 + b^2*d^4) + (4*A*a*b^2*d^2*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2

*d^5)) + (a*b^2*((A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)) - (-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)))^(1/2)

*(a + b*tan(c + d*x))^(1/2)*(-16*A^4*b^2*d^4)^(1/2)*8i)/(16*A^3*b^4*d + 16*A^3*a^2*b^2*d - (16*A^3*a^2*b^4*d^5

)/(a^2*d^4 + b^2*d^4) - (16*A^3*a^4*b^2*d^5)/(a^2*d^4 + b^2*d^4) + (4*A*a^3*b^2*d^4*(-16*A^4*b^2*d^4)^(1/2))/(

a^2*d^5 + b^2*d^5) + (4*A*a*b^4*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (A^2*a^2*b^2*d^2*((A^2*a*d

^2)/(4*(a^2*d^4 + b^2*d^4)) - (-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/

2)*32i)/(16*A^3*b^4*d + 16*A^3*a^2*b^2*d - (16*A^3*a^2*b^4*d^5)/(a^2*d^4 + b^2*d^4) - (16*A^3*a^4*b^2*d^5)/(a^

2*d^4 + b^2*d^4) + (4*A*a^3*b^2*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (4*A*a*b^4*d^4*(-16*A^4*b^2

*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)))*((A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)) - (-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4

+ b^2*d^4)))^(1/2)*2i - atan((a*b^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) + (A^2*a*d^2)/(4*(a^2*d

^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-16*A^4*b^2*d^4)^(1/2)*8i)/((16*A^3*a^2*b^4*d^5)/(a^2*d^4 +

b^2*d^4) - 16*A^3*a^2*b^2*d - 16*A^3*b^4*d + (16*A^3*a^4*b^2*d^5)/(a^2*d^4 + b^2*d^4) + (4*A*a^3*b^2*d^4*(-16*

A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (4*A*a*b^4*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (A^2*

b^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) + (A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(

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

d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) + (A^2*a^2*b^2*d^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) + (A^2*a

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

- 16*A^3*a^2*b^2*d - 16*A^3*b^4*d + (16*A^3*a^4*b^2*d^5)/(a^2*d^4 + b^2*d^4) + (4*A*a^3*b^2*d^4*(-16*A^4*b^2*

d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (4*A*a*b^4*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)))*((-16*A^4*b^2*

d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) + (A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*2i - ((2*B*(a^2 + b^2))/(b^3*

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

*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-16*B^4*b^2*d^4)^(1/2)*8i)/((1

6*B^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4) + (4*B*b^5*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3

*b^3*d^5)/(a^2*d^4 + b^2*d^4) + (4*B*a^2*b^3*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (B^2*b^2*(- (

-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*

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

d^5)) + (B^2*a^2*b^2*d^2*(- (-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d

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

2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3*b^3*d^5)/(a^2*d^4 + b^2*d^4) + (4*B*a^2*b^3*d^4*(-16*B^4*b^2*d

^4)^(1/2))/(a^2*d^5 + b^2*d^5)))*(- (-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4

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

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

\(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [343]

Optimal result