\(\int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx\) [198]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\frac {4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}-\frac {2 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d} \]

[Out]

-2/5*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b/d+4/5*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2
*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b/cos(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2648, 2721, 2719} \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\frac {4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b d} \]

[In]

Int[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^2,x]

[Out]

(4*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]) - (2*(d*Cos[a + b*x])^(3/2)*Sin[a
+ b*x])/(5*b*d)

Rule 2648

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d}+\frac {2}{5} \int \sqrt {d \cos (a+b x)} \, dx \\ & = -\frac {2 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d}+\frac {\left (2 \sqrt {d \cos (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \, dx}{5 \sqrt {\cos (a+b x)}} \\ & = \frac {4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}-\frac {2 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\frac {d \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \sin ^3(a+b x)}{3 b \sqrt {d \cos (a+b x)}} \]

[In]

Integrate[Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^2,x]

[Out]

(d*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 3/2, 5/2, Sin[a + b*x]^2]*Sin[a + b*x]^3)/(3*b*Sqrt[d*Cos[a +
b*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(193\) vs. \(2(85)=170\).

Time = 0.50 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.81

method	result	size
default	\(\frac {4 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d \left (4 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-8 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+5 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-\cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(194\)

[In]

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

[Out]

4/5*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d*(4*cos(1/2*b*x+1/2*a)^7-8*cos(1/2*b*x+1/2*a)^5
+5*cos(1/2*b*x+1/2*a)^3+(sin(1/2*b*x+1/2*a)^2)^(1/2)*(1-2*cos(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/
2*a),2^(1/2))-cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/sin(1/2*b*x+1/2*a)/
(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - i \, \sqrt {2} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + i \, \sqrt {2} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )\right )}}{5 \, b} \]

[In]

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

[Out]

-2/5*(sqrt(d*cos(b*x + a))*cos(b*x + a)*sin(b*x + a) - I*sqrt(2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPIn
verse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + I*sqrt(2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4
, 0, cos(b*x + a) - I*sin(b*x + a))))/b

Sympy [F]

\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int \sqrt {d \cos {\left (a + b x \right )}} \sin ^{2}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {d \cos (a+b x)} \sin ^2(a+b x) \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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