3.45 \(\int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=114 \[ \frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d \sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e} \]

[Out]

-2/3*(3*a^2+2*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/
2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)+10/3*a*b*(e*sin(d*x+c))^(1/2)/d/e+2/3*b*(a+b*cos(d*x+c
))*(e*sin(d*x+c))^(1/2)/d/e

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2692, 2669, 2642, 2641} \[ \frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d \sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*a^2 + 2*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) + (10*a*b*S
qrt[e*Sin[c + d*x]])/(3*d*e) + (2*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e)

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx &=\frac {2 b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2}{3} \int \frac {\frac {3 a^2}{2}+b^2+\frac {5}{2} a b \cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e}+\frac {1}{3} \left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx\\ &=\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e}+\frac {\left (\left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 \left (3 a^2+2 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.38, size = 79, normalized size = 0.69 \[ \frac {2 b \sin (c+d x) (6 a+b \cos (c+d x))-2 \left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )}{3 d \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a^2 + 2*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sqrt[Sin[c + d*x]] + 2*b*(6*a + b*Cos[c + d*x])*Sin[c
+ d*x])/(3*d*Sqrt[e*Sin[c + d*x]])

________________________________________________________________________________________

fricas [F]  time = 1.17, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {e \sin \left (d x + c\right )}}{e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(e*sin(d*x + c))/(e*sin(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c) + a)^2/sqrt(e*sin(d*x + c)), x)

________________________________________________________________________________________

maple [A]  time = 0.22, size = 170, normalized size = 1.49 \[ -\frac {3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-2 b^{2} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-12 a b \sin \left (d x +c \right ) \cos \left (d x +c \right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Elliptic
F((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip
ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-2*b^2*sin(d*x+c)*cos(d*x+c)^2-12*a*b*sin(d*x+c)*cos(d*x+c))/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c) + a)^2/sqrt(e*sin(d*x + c)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{\sqrt {e\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{2}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________