\(\int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx\) [610]

Optimal result

Integrand size = 25, antiderivative size = 151 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.25 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\frac {\sqrt [4]{-1} \left (\sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+\frac {2 \sqrt {a} \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \] Input:

Integrate[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

((-1)^(1/4)*(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]]]) + (2*Sqrt[a]*Sqrt[ 
b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x]) 
/a])/Sqrt[a + b*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4058, 609, 65, 219, 2035, 2257, 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.

\(\displaystyle \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4058

\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 609

\(\displaystyle \frac {b \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {2 b \int \frac {1}{1-\frac {b \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \int \frac {b-a \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \int \left (\frac {i (b-i a)}{2 (i-\tan (c+d x)) \sqrt {a+b \tan (c+d x)}}-\frac {i (-i a-b)}{2 (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \left (\frac {1}{2} \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\)

Input:

Int[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

(2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] 
- 2*(-1/2*(Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
+ b*Tan[c + d*x]]]) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d 
*x]])/Sqrt[a + b*Tan[c + d*x]]])/2))/d
 

Defintions of rubi rules used

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 609

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S 
ymbol] :> Simp[d*(e/b)   Int[(e*x)^(m - 1)*(c + d*x)^(n - 1), x], x] - Simp 
[e/b   Int[(e*x)^(m - 1)*(c + d*x)^(n - 1)*((a*d - b*c*x)/(a + b*x^2)), x], 
 x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[0, n, 1] && LtQ[0, m, 1] &&  !Integ 
erQ[m] &&  !IntegerQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4058

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]
 

Maple [B] (warning: unable to verify)

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

Time = 1.08 (sec) , antiderivative size = 1088397, normalized size of antiderivative = 7207.93

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

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2724 vs. \(2 (119) = 238\).

Time = 0.41 (sec) , antiderivative size = 5450, normalized size of antiderivative = 36.09 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\tan {\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(sqrt(a + b*tan(c + d*x))*sqrt(tan(c + d*x)), x)
 

Maxima [F]

\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \sqrt {\tan \left (d x + c\right )} \,d x } \] Input:

integrate(tan(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(tan(d*x + c)), x)
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 4.33 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.91 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\frac {4\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}-\sqrt {a}}\right )}{d}-\mathrm {atan}\left (\frac {2\,\left (d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {-b+a\,1{}\mathrm {i}}{4\,d^2}}-\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {-b+a\,1{}\mathrm {i}}{4\,d^2}}\right )}{a\,1{}\mathrm {i}+b\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}\right )\,\sqrt {-\frac {-b+a\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {2\,\left (\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {b+a\,1{}\mathrm {i}}{4\,d^2}}-d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {b+a\,1{}\mathrm {i}}{4\,d^2}}\right )}{a\,1{}\mathrm {i}+b\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}\right )\,\sqrt {\frac {b+a\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan((2*(a^(1/2)*d*tan(c + d*x)^(1/2)*((a*1i + b)/(4*d^2))^(1/2) - d*tan(c 
 + d*x)^(1/2)*(a + b*tan(c + d*x))^(1/2)*((a*1i + b)/(4*d^2))^(1/2)))/(a*1 
i + b*tan(c + d*x)*1i - a^(1/2)*(a + b*tan(c + d*x))^(1/2)*1i))*((a*1i + b 
)/(4*d^2))^(1/2)*2i - atan((2*(d*tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^( 
1/2)*(-(a*1i - b)/(4*d^2))^(1/2) - a^(1/2)*d*tan(c + d*x)^(1/2)*(-(a*1i - 
b)/(4*d^2))^(1/2)))/(a*1i + b*tan(c + d*x)*1i - a^(1/2)*(a + b*tan(c + d*x 
))^(1/2)*1i))*(-(a*1i - b)/(4*d^2))^(1/2)*2i + (4*b^(1/2)*atanh((b^(1/2)*t 
an(c + d*x)^(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))))/d
 

Reduce [F]

\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}d x \] Input:

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

Output:

int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a),x)