\(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [434]

Optimal result

Integrand size = 35, antiderivative size = 323 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \]

[Out]

(I*a-b)^(3/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+1/8*(6*A*a^2*b-16*A*b^3-
B*a^3-24*B*a*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d+(I*a+b)^(3/2)*(A-I*B)*arc
tanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+1/8*(6*A*a*b-B*a^2-8*B*b^2)*tan(d*x+c)^(1/2)*(a+
b*tan(d*x+c))^(1/2)/b/d+1/12*(6*A*b-B*a)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2)/b/d+1/3*B*tan(d*x+c)^(1/2)*(a
+b*tan(d*x+c))^(5/2)/b/d

Rubi [A] (verified)

Time = 3.54 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3688, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {\left (a^3 (-B)+6 a^2 A b-24 a b^2 B-16 A b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(-b+i a)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(b+i a)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \]

[In]

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

[Out]

((I*a - b)^(3/2)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((6*a^2*A*
b - 16*A*b^3 - a^3*B - 24*a*b^2*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(8*b^(3/2)*
d) + ((I*a + b)^(3/2)*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((6*
a*A*b - a^2*B - 8*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(8*b*d) + ((6*A*b - a*B)*Sqrt[Tan[c + d*
x]]*(a + b*Tan[c + d*x])^(3/2))/(12*b*d) + (B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2))/(3*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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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 212

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

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 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 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 3728

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
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {(a+b \tan (c+d x))^{3/2} \left (-\frac {a B}{2}-3 b B \tan (c+d x)+\frac {1}{2} (6 A b-a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 b} \\ & = \frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {3}{4} a (2 A b+a B)-6 b (A b+a B) \tan (c+d x)+\frac {3}{4} \left (6 a A b-a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 b} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\int \frac {-\frac {3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{6 b} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\text {Subst}\left (\int \frac {-\frac {3}{8} a \left (10 a A b+a^2 B-8 b^2 B\right )-6 b \left (2 a A b+a^2 B-b^2 B\right ) x+\frac {3}{8} \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}+\frac {\text {Subst}\left (\int \left (\frac {3 \left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b d} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\text {Subst}\left (\int \frac {b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{16 b d} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\text {Subst}\left (\int \left (\frac {i b \left (a^2 A-A b^2-2 a b B\right )-b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i b \left (a^2 A-A b^2-2 a b B\right )+b \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 b d} \\ & = \frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\left ((a+i b)^2 (i A-B)\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left ((a-i b)^2 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b d} \\ & = \frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\left ((a+i b)^2 (i A-B)\right ) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left ((a-i b)^2 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \\ & = \frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}+\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.65 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.07 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {24 \sqrt [4]{-1} (-a+i b)^{3/2} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 (-1)^{3/4} (a+i b)^{3/2} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-3 \left (-6 a A b+a^2 B+8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 (6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}-\frac {3 \sqrt {a} \left (-6 a^2 A b+16 A b^3+a^3 B+24 a b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{24 b d} \]

[In]

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

[Out]

(24*(-1)^(1/4)*(-a + I*b)^(3/2)*b*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T
an[c + d*x]]] + 24*(-1)^(3/4)*(a + I*b)^(3/2)*b*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])
/Sqrt[a + b*Tan[c + d*x]]] - 3*(-6*a*A*b + a^2*B + 8*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*(6
*A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2) + 8*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2)
- (3*Sqrt[a]*(-6*a^2*A*b + 16*A*b^3 + a^3*B + 24*a*b^2*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1
+ (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/(24*b*d)

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.90 (sec) , antiderivative size = 2403184, normalized size of antiderivative = 7440.20

\[\text {output too large to display}\]

[In]

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

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12212 vs. \(2 (264) = 528\).

Time = 5.20 (sec) , antiderivative size = 24426, normalized size of antiderivative = 75.62 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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