Integrand size = 25, antiderivative size = 155 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=-\frac {i \sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {i \sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \] Output:
-I*(I*a-b)^(1/2)*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*(I*a+b)^(1/2)*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
Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\frac {(-1)^{3/4} \left (\sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \] Input:
Integrate[Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]],x]
Output:
((-1)^(3/4)*(Sqrt[-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]*ArcTan[((-1)^(1/4)*Sqrt[ a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])*Sqrt[Cot[c + d*x]] *Sqrt[Tan[c + d*x]])/d
Time = 0.44 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4729, 3042, 4058, 610, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 610 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \left (\frac {i a-b}{2 (i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {i a+b}{2 \sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-i \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-i \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\) |
Input:
Int[Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]],x]
Output:
(((-I)*Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* Tan[c + d*x]]] - I*Sqrt[I*a + 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]])/d
Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S ymbol] :> Simp[e^(m + 1/2) Int[ExpandIntegrand[1/(Sqrt[e*x]*Sqrt[c + d*x] ), x^(m + 1/2)*((c + d*x)^(n + 1/2)/(a + b*x^2)), x], x], x] /; FreeQ[{a, b , c, d, e}, x] && IGtQ[n + 1/2, 0] && ILtQ[m - 1/2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1009\) vs. \(2(125)=250\).
Time = 5.80 (sec) , antiderivative size = 1010, normalized size of antiderivative = 6.52
Input:
int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4/d*2^(1/2)/a/(-b+(a^2+b^2)^(1/2))^(1/2)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c ))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2)*ln(1/(1-cos(d*x+c))*(- a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*2^(1/2)*( (a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2 )^(1/2))^(1/2)*sin(d*x+c)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*(-b+(a^2+b^2)^ (1/2))^(1/2)-(b+(a^2+b^2)^(1/2))^(1/2)*(a^2+b^2)^(1/2)*ln(1/(1-cos(d*x+c)) *(a*(1-cos(d*x+c))^2*csc(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin (d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*(a^ 2+b^2)^(1/2)*(1-cos(d*x+c))-2*b*(1-cos(d*x+c))-sin(d*x+c)*a))*(-b+(a^2+b^2 )^(1/2))^(1/2)-(b+(a^2+b^2)^(1/2))^(1/2)*b*ln(1/(1-cos(d*x+c))*(-a*(1-cos( d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*2^(1/2)*((a*cos(d* x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(b+(a^2+b^2)^(1/2))^ (1/2)*sin(d*x+c)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*(-b+(a^2+b^2)^(1/2))^(1 /2)+(b+(a^2+b^2)^(1/2))^(1/2)*b*ln(1/(1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*cs c(d*x+c)+2*2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^ 2)^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*(a^2+b^2)^(1/2)*(1-cos(d*x +c))-2*b*(1-cos(d*x+c))-sin(d*x+c)*a))*(-b+(a^2+b^2)^(1/2))^(1/2)-2*arctan (1/(-b+(a^2+b^2)^(1/2))^(1/2)*((b+(a^2+b^2)^(1/2))^(1/2)*(-cot(d*x+c)+csc( d*x+c))+2^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^ (1/2))/(1-cos(d*x+c))*sin(d*x+c))*a^2-2*arctan(1/(-b+(a^2+b^2)^(1/2))^(...
Leaf count of result is larger than twice the leaf count of optimal. 2765 vs. \(2 (119) = 238\).
Time = 0.30 (sec) , antiderivative size = 2765, normalized size of antiderivative = 17.84 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-1/8*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((((a^4*b + 4*a^2*b^3)*d*tan(d *x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c) - (3*a^4*b + 4*a^ 2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^ 3*b + 2*a*b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2) + 2*((a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b + 2* a^2*b^3)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c)^2 - (a^4 + 3 *a^2*b^2 + 4*b^4)*d^2*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)/sqrt(tan(d*x + c)))/(tan(d*x + c)^2 + 1)) - 1/8*sqrt(-(d^2*sqrt(-a^2/d^ 4) + b)/d^2)*log(-(((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3* b^2 + 4*a*b^4)*d*tan(d*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2) + 2*((a^5 + 3*a^ 3*b^2 + 4*a*b^4)*tan(d*x + c)^2 + 2*(a^4*b + 2*a^2*b^3)*tan(d*x + c) + (2* (a^3*b + 2*a*b^3)*d^2*tan(d*x + c)^2 - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2*tan(d *x + c))*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)/sqrt(tan(d*x + c)))/(tan (d*x + c)^2 + 1)) + 1/8*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c ) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan (d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(- (d^2*sqrt(-a^2/d^4) + b)/d^2) - 2*((a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x ...
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\cot {\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**(1/2)*(a+b*tan(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a + b*tan(c + d*x))*sqrt(cot(c + d*x)), x)
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \sqrt {\cot \left (d x + c\right )} \,d x } \] Input:
integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*tan(d*x + c) + a)*sqrt(cot(d*x + c)), x)
Timed out. \[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \] Input:
int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(1/2),x)
Output:
int(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(1/2), x)
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a +\tan \left (d x +c \right ) b}\, \sqrt {\cot \left (d x +c \right )}d x \] Input:
int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2),x)
Output:
int(sqrt(tan(c + d*x)*b + a)*sqrt(cot(c + d*x)),x)