Integrand size = 33, antiderivative size = 118 \[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\frac {2 \sqrt {\cos (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{e \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}} \] Output:
2*cos(e*x+d)^(1/2)*EllipticE(sin(1/2*d+1/2*e*x-1/2*arctan(c,a)),2^(1/2)*(( a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*(a+b*sec(e*x+d)+c*tan(e*x+d))^( 1/2)/e/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)
Result contains complex when optimal does not.
Time = 31.90 (sec) , antiderivative size = 54676, normalized size of antiderivative = 463.36 \[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\text {Result too large to show} \] Input:
Integrate[Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]],x]
Output:
Result too large to show
Time = 0.52 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3642, 3042, 3598, 3042, 3132}
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.
\(\displaystyle \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}dx\) |
\(\Big \downarrow \) 3642 |
\(\displaystyle \frac {\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{\sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{\sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\) |
\(\Big \downarrow \) 3598 |
\(\displaystyle \frac {\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {2 \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
Input:
Int[Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]],x]
Output:
(2*Sqrt[Cos[d + e*x]]*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(e *Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ )]], x_Symbol] :> Simp[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])] Int[Sqrt[a/(a + S qrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - Arc Tan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Int[cos[(d_.) + (e_.)*(x_)]^(n_)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + ( c_.)*tan[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[d + e*x]^n*((a + b*Sec[d + e*x] + c*Tan[d + e*x])^n/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n) Int[(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d , e}, x] && !IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 8.16 (sec) , antiderivative size = 2660, normalized size of antiderivative = 22.54
method | result | size |
risch | \(\text {Expression too large to display}\) | \(2660\) |
default | \(\text {Expression too large to display}\) | \(12633\) |
Input:
int(cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2),x,method=_RETURNV ERBOSE)
Output:
I*(-I*exp(I*(e*x+d))^2*c+a*exp(I*(e*x+d))^2+I*c+2*b*exp(I*(e*x+d))+a)/e/(e xp(I*(e*x+d))*(-I*exp(I*(e*x+d))^2*c+a*exp(I*(e*x+d))^2+I*c+2*b*exp(I*(e*x +d))+a))^(1/2)*(1/exp(RootOf(_Z^2+1,index=1)*(e*x+d))*(exp(RootOf(_Z^2+1,i ndex=1)*(e*x+d))^2+1))^(1/2)*(-(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,i ndex=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-RootOf(_Z^2+1 ,index=1)*c-2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))-a)/(exp(RootOf(_Z^2+1, index=1)*(e*x+d))^2+1))^(1/2)*(-(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1, index=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-RootOf(_Z^2+ 1,index=1)*c-2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))-a)*(exp(RootOf(_Z^2+1 ,index=1)*(e*x+d))^2+1))^(1/2)/(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,i ndex=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-RootOf(_Z^2+1 ,index=1)*c-2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))-a)*(exp(RootOf(_Z^2+1, index=1)*(e*x+d))*(-RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,index=1)*(e*x +d))^2*c+exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a+RootOf(_Z^2+1,index=1)*c+ 2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))+a))^(1/2)/((-RootOf(_Z^2+1,index=1 )*exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*c+exp(RootOf(_Z^2+1,index=1)*(e*x+ d))^2*a+RootOf(_Z^2+1,index=1)*c+2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))+a )*(exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2+1))^(1/2)*2^(1/2)-I/e*(-2*b*(-b+( -a^2+b^2-c^2)^(1/2))/(I*c-a)*((exp(I*(e*x+d))+(-b+(-a^2+b^2-c^2)^(1/2))/(I *c-a))/(-b+(-a^2+b^2-c^2)^(1/2))*(I*c-a))^(1/2)*((exp(I*(e*x+d))-(b+(-a...
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 1369, normalized size of antiderivative = 11.60 \[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2),x, algorith m="fricas")
Output:
-2/3*((I*a*b - b*c)*sqrt(1/2*a - 1/2*I*c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^ 4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9*a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8* a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b + 2*I*b*c + 3* (a^2 + c^2)*cos(e*x + d) - 3*(-I*a^2 - I*c^2)*sin(e*x + d))/(a^2 + c^2)) + (-I*a*b - b*c)*sqrt(1/2*a + 1/2*I*c)*weierstrassPInverse(-4/3*(3*a^4 - 4* a^2*b^2 + 4*b^2*c^2 - 6*I*a*c^3 - 3*c^4 - 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 + 9*I*b*c^5 - 2*I *(9*a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 - 3*I*(9*a^4*b - 8*a^2* b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b - 2*I*b*c + 3*(a^2 + c^2)*cos(e*x + d) - 3*(I*a^2 + I*c^2)*sin(e*x + d))/(a^2 + c^2)) + 3*(- I*a^2 - I*c^2)*sqrt(1/2*a - 1/2*I*c)*weierstrassZeta(-4/3*(3*a^4 - 4*a^2*b ^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2 *c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9*a ^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2*b^3)* c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2 *I*(9*a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8...
\[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\int \sqrt {a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}} \sqrt {\cos {\left (d + e x \right )}}\, dx \] Input:
integrate(cos(e*x+d)**(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))**(1/2),x)
Output:
Integral(sqrt(a + b*sec(d + e*x) + c*tan(d + e*x))*sqrt(cos(d + e*x)), x)
\[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\int { \sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )} \,d x } \] Input:
integrate(cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2),x, algorith m="maxima")
Output:
integrate(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)*sqrt(cos(e*x + d)), x)
\[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\int { \sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )} \,d x } \] Input:
integrate(cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2),x, algorith m="giac")
Output:
integrate(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)*sqrt(cos(e*x + d)), x)
Timed out. \[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\int \sqrt {\cos \left (d+e\,x\right )}\,\sqrt {a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}} \,d x \] Input:
int(cos(d + e*x)^(1/2)*(a + c*tan(d + e*x) + b/cos(d + e*x))^(1/2),x)
Output:
int(cos(d + e*x)^(1/2)*(a + c*tan(d + e*x) + b/cos(d + e*x))^(1/2), x)
\[ \int \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \, dx=\int \sqrt {\sec \left (e x +d \right ) b +\tan \left (e x +d \right ) c +a}\, \sqrt {\cos \left (e x +d \right )}d x \] Input:
int(cos(e*x+d)^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2),x)
Output:
int(sqrt(sec(d + e*x)*b + tan(d + e*x)*c + a)*sqrt(cos(d + e*x)),x)