3.1.46 \(\int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx\) [46]

Optimal. Leaf size=118 \[ -\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt {e \sin (c+d x)}}-\frac {2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a b (e \sin (c+d x))^{3/2}}{d e^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 118, 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 = {2770, 2748, 2721, 2719} \begin {gather*} -\frac {2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{d e \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(3/2),x]

[Out]

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

Rule 2719

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 2770

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

Rubi steps

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

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Mathematica [A]
time = 0.26, size = 75, normalized size = 0.64 \begin {gather*} \frac {-4 a b-2 \left (a^2+b^2\right ) \cos (c+d x)+2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sqrt {\sin (c+d x)}}{d e \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(3/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(282\) vs. \(2(140)=280\).
time = 0.12, size = 283, normalized size = 2.40

method	result	size
default	\(\frac {2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-2 a^{2} \left (\cos ^{2}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}-4 \cos \left (d x +c \right ) a b}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE
((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+4*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipt
icE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^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))*a^2-2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*El
lipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-2*a^2*cos(d*x+c)^2-2*cos(d*x+c)^2*b^2-4*cos(d*x+c)*a*b)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

e^(-3/2)*integrate((b*cos(d*x + c) + a)^2/sin(d*x + c)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 134, normalized size = 1.14 \begin {gather*} \frac {{\left (\sqrt {2} \sqrt {-i} {\left (-i \, a^{2} - 2 i \, b^{2}\right )} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} \sqrt {i} {\left (i \, a^{2} + 2 i \, b^{2}\right )} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\sin \left (d x + c\right )}\right )} e^{\left (-\frac {3}{2}\right )}}{d \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

(sqrt(2)*sqrt(-I)*(-I*a^2 - 2*I*b^2)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c)
+ I*sin(d*x + c))) + sqrt(2)*sqrt(I)*(I*a^2 + 2*I*b^2)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse
(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2*a*b + (a^2 + b^2)*cos(d*x + c))*sqrt(sin(d*x + c)))*e^(-3/2)/(d*
sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{2}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2/(e*sin(d*x+c))**(3/2),x)

[Out]

Integral((a + b*cos(c + d*x))**2/(e*sin(c + d*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2*e^(-3/2)/sin(d*x + c)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^2/(e*sin(c + d*x))^(3/2),x)

[Out]

int((a + b*cos(c + d*x))^2/(e*sin(c + d*x))^(3/2), x)

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