Integrand size = 32, antiderivative size = 160 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{2 \sqrt {2} \left (b^2+c^2\right )^{3/4} e}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \] Output:
1/4*arctanh(1/2*(b^2+c^2)^(1/4)*sin(d+e*x-arctan(c,b))*2^(1/2)/((b^2+c^2)^ (1/2)+(b^2+c^2)^(1/2)*cos(d+e*x-arctan(c,b)))^(1/2))*2^(1/2)/(b^2+c^2)^(3/ 4)/e-1/2*(c*cos(e*x+d)-b*sin(e*x+d))/(b^2+c^2)^(1/2)/e/((b^2+c^2)^(1/2)+b* cos(e*x+d)+c*sin(e*x+d))^(3/2)
Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\text {\$Aborted} \] Input:
Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-3/2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 3595, 3042, 3594, 3042, 3128, 219}
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.
\(\displaystyle \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3595 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \cos (d+e x)+c \sin (d+e x)+\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3594 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )+\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)+\frac {\pi }{2}\right )+\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle -\frac {\int \frac {1}{2 \sqrt {b^2+c^2}-\frac {\left (b^2+c^2\right ) \sin ^2\left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )+\sqrt {b^2+c^2}}}d\left (-\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )+\sqrt {b^2+c^2}}}\right )}{2 e \sqrt {b^2+c^2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )+\sqrt {b^2+c^2}}}\right )}{2 \sqrt {2} e \left (b^2+c^2\right )^{3/4}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\) |
Input:
Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-3/2),x]
Output:
ArcTanh[((b^2 + c^2)^(1/4)*Sin[d + e*x - ArcTan[b, c]])/(Sqrt[2]*Sqrt[Sqrt [b^2 + c^2] + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]])]/(2*Sqrt[2]*(b ^2 + c^2)^(3/4)*e) - (c*Cos[d + e*x] - b*Sin[d + e*x])/(2*Sqrt[b^2 + c^2]* e*(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(3/2))
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])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( x_)]], x_Symbol] :> Int[1/Sqrt[a + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (n_), x_Symbol] :> Simp[(c*Cos[d + e*x] - b*Sin[d + e*x])*((a + b*Cos[d + e *x] + c*Sin[d + e*x])^n/(a*e*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(137)=274\).
Time = 0.73 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.19
method | result | size |
default | \(-\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}+2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {7}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+b^{2}+c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e}\) | \(350\) |
Input:
int(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x,method=_RETURNVE RBOSE)
Output:
-1/4/(b^2+c^2)^(7/4)*(sin(e*x+d-arctan(-b,c))*2^(1/2)*arctanh(1/2*(-(b^2+c ^2)^(1/2)*sin(e*x+d-arctan(-b,c))+(b^2+c^2)^(1/2))^(1/2)*2^(1/2)/(b^2+c^2) ^(1/4))*(b^2+c^2)+2^(1/2)*arctanh(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(- b,c))+(b^2+c^2)^(1/2))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))*b^2+2^(1/2)*arctanh( 1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+(b^2+c^2)^(1/2))^(1/2)*2^(1/ 2)/(b^2+c^2)^(1/4))*c^2+2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+(b^2+c ^2)^(1/2))^(1/2)*(b^2+c^2)^(3/4))*(-(b^2+c^2)^(1/2)*(sin(e*x+d-arctan(-b,c ))-1))^(1/2)/cos(e*x+d-arctan(-b,c))/((b^2*sin(e*x+d-arctan(-b,c))+c^2*sin (e*x+d-arctan(-b,c))+b^2+c^2)/(b^2+c^2)^(1/2))^(1/2)/e
Leaf count of result is larger than twice the leaf count of optimal. 622 vs. \(2 (143) = 286\).
Time = 0.25 (sec) , antiderivative size = 622, normalized size of antiderivative = 3.89 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
integrate(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm ="fricas")
Output:
-1/4*(sqrt(1/2)*(3*b^2*c*cos(e*x + d) - (3*b^2*c - c^3)*cos(e*x + d)^3 - ( b^3 - (b^3 - 3*b*c^2)*cos(e*x + d)^2)*sin(e*x + d))*(b^2 + c^2)^(1/4)*log( ((3*b^2*c - c^3)*cos(e*x + d)^3 - 4*sqrt(1/2)*(2*b*c*cos(e*x + d)*sin(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + b^2 + 2*c^2 - 2*sqrt(b^2 + c^2)*(b*cos (e*x + d) + c*sin(e*x + d)))*(b^2 + c^2)^(1/4)*sqrt(b*cos(e*x + d) + c*sin (e*x + d) + sqrt(b^2 + c^2)) + (b^2*c + 4*c^3)*cos(e*x + d) - (3*b^3 + 4*b *c^2 + (b^3 - 3*b*c^2)*cos(e*x + d)^2)*sin(e*x + d) - 4*(2*b*c*cos(e*x + d )^2 - (b^2 - c^2)*cos(e*x + d)*sin(e*x + d) - b*c)*sqrt(b^2 + c^2))/(3*b^2 *c*cos(e*x + d) - (3*b^2*c - c^3)*cos(e*x + d)^3 - (b^3 - (b^3 - 3*b*c^2)* cos(e*x + d)^2)*sin(e*x + d))) + 2*(2*(b^3 + b*c^2)*cos(e*x + d) + 2*(b^2* c + c^3)*sin(e*x + d) - (2*b*c*cos(e*x + d)*sin(e*x + d) + (b^2 - c^2)*cos (e*x + d)^2 + b^2 + 2*c^2)*sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e* x + d) + sqrt(b^2 + c^2)))/((3*b^4*c + 2*b^2*c^3 - c^5)*e*cos(e*x + d)^3 - 3*(b^4*c + b^2*c^3)*e*cos(e*x + d) - ((b^5 - 2*b^3*c^2 - 3*b*c^4)*e*cos(e *x + d)^2 - (b^5 + b^3*c^2)*e)*sin(e*x + d))
\[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )} + \sqrt {b^{2} + c^{2}}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/((b**2+c**2)**(1/2)+b*cos(e*x+d)+c*sin(e*x+d))**(3/2),x)
Output:
Integral((b*cos(d + e*x) + c*sin(d + e*x) + sqrt(b**2 + c**2))**(-3/2), x)
Exception generated. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm ="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x, algorithm ="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )+\sqrt {b^2+c^2}\right )}^{3/2}} \,d x \] Input:
int(1/(b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^(3/2),x)
Output:
int(1/(b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^(3/2), x)
\[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\int \frac {1}{\left (\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right )^{\frac {3}{2}}}d x \] Input:
int(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x)
Output:
int(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x)