\(\int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\) [645]

Optimal result

Integrand size = 35, antiderivative size = 210 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 (A b-a B)}{\left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4326, 3689, 3697, 3696, 95, 209, 212} \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {2 (A b-a B)}{d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {(A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac {(A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

[In]

Int[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)),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]])/((I*a - 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]])/((I*a + b)^(3/2)*d) - (2*(A*b - a*B))/((a^2 + b^2)*d*Sqrt[Cot[c + d*
x]]*Sqrt[a + b*Tan[c + d*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 209

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

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 3689

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f
*(m + 1)*(a^2 + b^2))), x] + Dist[1/(b*(m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f
*x])^(n - 1)*Simp[b*B*(b*c*(m + 1) + a*d*n) + A*b*(a*c*(m + 1) - b*d*n) - b*(A*(b*c - a*d) - B*(a*c + b*d))*(m
+ 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B},
x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegerQ[
m] || IntegersQ[2*m, 2*n])

Rule 3696

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

Rule 3697

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

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

Mathematica [A] (verified)

Time = 2.35 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\frac {\sqrt [4]{-1} a (a+i b) (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} a (a-i b) (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}}+2 (-A b+a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}\right )}{a \left (a^2+b^2\right ) d} \]

[In]

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

[Out]

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

Maple [B] (warning: unable to verify)

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

Time = 1.32 (sec) , antiderivative size = 1559507, normalized size of antiderivative = 7426.22

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

[In]

int((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),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. 18672 vs. \(2 (170) = 340\).

Time = 7.01 (sec) , antiderivative size = 18672, normalized size of antiderivative = 88.91 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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