\(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [641]

Optimal result

Integrand size = 35, antiderivative size = 266 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}+\frac {(2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{b^{3/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d}+\frac {B \sqrt {a+b \tan (c+d x)}}{b d \sqrt {\cot (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.78 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4326, 3688, 3736, 6857, 65, 223, 212, 95, 211, 214} \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=-\frac {(A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(2 A b-a B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {(A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}+\frac {B \sqrt {a+b \tan (c+d x)}}{b d \sqrt {\cot (c+d x)}} \]

[In]

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

[Out]

-(((A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c
+ d*x]])/(Sqrt[I*a - b]*d)) + ((2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*S
qrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(b^(3/2)*d) - ((A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt
[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[I*a + b]*d) + (B*Sqrt[a + b*Tan[c + d*x]])/
(b*d*Sqrt[Cot[c + d*x]])

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 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 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

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

Mathematica [A] (verified)

Time = 2.69 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\sqrt {a} \sqrt {-a+i b} \sqrt {a+i b} (-2 A b+a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (\sqrt [4]{-1} \sqrt {a+i b} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {-a+i b} \left ((-1)^{3/4} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {a+i b} B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))\right )\right )\right )}{\sqrt {-a+i b} \sqrt {a+i b} b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-(Sqrt[a]*Sqrt[-a + I*b]*Sqrt[a + I*b]*(-2*A*b + a*B)*ArcSinh[(Sqrt[b]
*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a]) + Sqrt[b]*((-1)^(1/4)*Sqrt[a + I*b]*b*(I*A + B)*Ar
cTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[
-a + I*b]*((-1)^(3/4)*b*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]
]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[a + I*b]*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])))))/(Sqrt[-a + I*b]*Sqrt
[a + I*b]*b^(3/2)*d*Sqrt[a + b*Tan[c + d*x]])

Maple [B] (warning: unable to verify)

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

Time = 1.25 (sec) , antiderivative size = 1890515, normalized size of antiderivative = 7107.20

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

[In]

int((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(1/2),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. 10269 vs. \(2 (214) = 428\).

Time = 5.04 (sec) , antiderivative size = 20571, normalized size of antiderivative = 77.33 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(c
onst gen &

Mupad [F(-1)]

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

[In]

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

[Out]

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