\(\int \frac {1}{(\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^{5/2}} \, dx\) [436]

Optimal result

Integrand size = 32, antiderivative size = 226 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{16 \sqrt {2} \left (b^2+c^2\right )^{5/4} e}-\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \]

[Out]

3/32*arctanh(1/2*(b^2+c^2)^(1/4)*sin(d+e*x-arctan(b,c))*2^(1/2)/((b^2+c^2)^(1/2)+cos(d+e*x-arctan(b,c))*(b^2+c
^2)^(1/2))^(1/2))/(b^2+c^2)^(5/4)/e*2^(1/2)+1/4*(-c*cos(e*x+d)+b*sin(e*x+d))/e/(b^2+c^2)^(1/2)/(b*cos(e*x+d)+c
*sin(e*x+d)+(b^2+c^2)^(1/2))^(5/2)-3/16*(c*cos(e*x+d)-b*sin(e*x+d))/(b^2+c^2)/e/(b*cos(e*x+d)+c*sin(e*x+d)+(b^
2+c^2)^(1/2))^(3/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3195, 3194, 2728, 212} \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )+\sqrt {b^2+c^2}}}\right )}{16 \sqrt {2} e \left (b^2+c^2\right )^{5/4}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 e \left (b^2+c^2\right ) \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \]

[In]

Int[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-5/2),x]

[Out]

(3*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]]])])/(16*Sqrt[2]*(b^2 + c^2)^(5/4)*e) - (c*Cos[d + e*x] - b*Sin[d + e*x])/(4*Sqrt[b^2
+ c^2]*e*(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)) - (3*(c*Cos[d + e*x] - b*Sin[d + e*x]))/(1
6*(b^2 + c^2)*e*(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(3/2))

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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3194

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]

Rule 3195

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}+\frac {3 \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx}{8 \sqrt {b^2+c^2}} \\ & = -\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {3 \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx}{32 \left (b^2+c^2\right )} \\ & = -\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}+\frac {3 \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}} \, dx}{32 \left (b^2+c^2\right )} \\ & = -\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {3 \text {Subst}\left (\int \frac {1}{2 \sqrt {b^2+c^2}-x^2} \, dx,x,-\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{16 \left (b^2+c^2\right ) e} \\ & = \frac {3 \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 )}{16 \sqrt {2} \left (b^2+c^2\right )^{5/4} e}-\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \\ \end{align*}

Mathematica [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {\$Aborted} \]

[In]

Integrate[(Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x])^(-5/2),x]

[Out]

$Aborted

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.55

method	result	size
default	\(\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}+2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {5}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+b^{2}+c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e}\)	\(350\)

[In]

int(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(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)/(b^2+c^2)^(5/4)/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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 895 vs. \(2 (203) = 406\).

Time = 0.50 (sec) , antiderivative size = 895, normalized size of antiderivative = 3.96 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^(5/2),x, algorithm="fricas")

[Out]

1/32*(3*sqrt(1/2)*(5*b^4*c*cos(e*x + d) + (5*b^4*c - 10*b^2*c^3 + c^5)*cos(e*x + d)^5 - 10*(b^4*c - b^2*c^3)*c
os(e*x + d)^3 - (b^5 + (b^5 - 10*b^3*c^2 + 5*b*c^4)*cos(e*x + d)^4 - 2*(b^5 - 5*b^3*c^2)*cos(e*x + d)^2)*sin(e
*x + d))*log(((3*b^2*c - c^3)*cos(e*x + d)^3 + (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*sqrt(1/2)*(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))/(b^2 + c^2)^(1/4) - 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)))/(b^2 + c^2)^(1/4) + 2*(3*(b^4 - 6*b^2*c^2 + c^4)*cos(e*x + d)^4 - 7*b^4
- 26*b^2*c^2 - 16*c^4 - 6*(2*b^4 - 3*b^2*c^2 - c^4)*cos(e*x + d)^2 + 12*((b^3*c - b*c^3)*cos(e*x + d)^3 - (2*b
^3*c + b*c^3)*cos(e*x + d))*sin(e*x + d) - 2*((b^3 - 3*b*c^2)*cos(e*x + d)^3 - 3*(3*b^3 + 2*b*c^2)*cos(e*x + d
) - (9*b^2*c + 8*c^3 - (3*b^2*c - c^3)*cos(e*x + d)^2)*sin(e*x + d))*sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*
sin(e*x + d) + sqrt(b^2 + c^2)))/((5*b^6*c - 5*b^4*c^3 - 9*b^2*c^5 + c^7)*e*cos(e*x + d)^5 - 10*(b^6*c - b^2*c
^5)*e*cos(e*x + d)^3 + 5*(b^6*c + b^4*c^3)*e*cos(e*x + d) - ((b^7 - 9*b^5*c^2 - 5*b^3*c^4 + 5*b*c^6)*e*cos(e*x
+ d)^4 - 2*(b^7 - 4*b^5*c^2 - 5*b^3*c^4)*e*cos(e*x + d)^2 + (b^7 + b^5*c^2)*e)*sin(e*x + d))

Sympy [F]

\[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )} + \sqrt {b^{2} + c^{2}}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b**2+c**2)**(1/2))**(5/2),x)

[Out]

Integral((b*cos(d + e*x) + c*sin(d + e*x) + sqrt(b**2 + c**2))**(-5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*cos(e*x+d)+c*sin(e*x+d)+(b^2+c^2)^(1/2))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )+\sqrt {b^2+c^2}\right )}^{5/2}} \,d x \]

[In]

int(1/(b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^(5/2),x)

[Out]

int(1/(b*cos(d + e*x) + c*sin(d + e*x) + (b^2 + c^2)^(1/2))^(5/2), x)