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

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{d e \sqrt {e \sin (c+d x)}}-\frac {2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a b (e \sin (c+d x))^{3/2}}{d e^3} \]

[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)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2770, 2748, 2721, 2719} \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a b (e \sin (c+d x))^{3/2}}{d e^3}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{d e \sqrt {e \sin (c+d x)}} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\frac {-4 a b-2 \left (a^2+b^2\right ) \cos (c+d x)+2 \left (a^2+2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sqrt {\sin (c+d x)}}{d e \sqrt {e \sin (c+d x)}} \]

[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]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(282\) vs. \(2(140)=280\).

Time = 3.48 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.40

method	result	size
default	\(\frac {2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+4 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-2 a^{2} \left (\cos ^{2}\left (d x +c \right )\right )-2 b^{2} \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right ) a b}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(283\)
parts	\(\frac {a^{2} \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b^{2} \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{2}\left (d x +c \right )\right )\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {4 a b}{\sqrt {e \sin \left (d x +c \right )}\, e d}\)	\(309\)

[In]

int((a+cos(d*x+c)*b)^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*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE(
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+4*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE
((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF(
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF
((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-2*a^2*cos(d*x+c)^2-2*b^2*cos(d*x+c)^2-4*cos(d*x+c)*a*b)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}}{d e^{2} \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[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)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[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*sin(d*x + c))^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[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)