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

Optimal. Leaf size=69 \[ \frac {4 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2568, 2642, 2641} \[ \frac {4 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2568

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.10, size = 58, normalized size = 0.84 \[ \frac {d \sin ^3(a+b x) \cos ^2(a+b x)^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {5}{2};\sin ^2(a+b x)\right )}{3 b (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 1\right )}}{d \cos \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 1)/(d*cos(b*x + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 188, normalized size = 2.72 \[ \frac {4 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4-(sin(
1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))-sin(1/2*b*x+1/2
*a)^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*co
s(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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