\(\int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) [216]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b d}-\frac {2 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2648, 2721, 2720} \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}-\frac {4 \sin (a+b x) \sqrt {d \cos (a+b x)}}{7 b d}+\frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b \sqrt {d \cos (a+b x)}} \]

[In]

Int[Sin[a + b*x]^4/Sqrt[d*Cos[a + b*x]],x]

[Out]

(8*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(7*b*Sqrt[d*Cos[a + b*x]]) - (4*Sqrt[d*Cos[a + b*x]]*Sin[a +
b*x])/(7*b*d) - (2*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^3)/(7*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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d}+\frac {6}{7} \int \frac {\sin ^2(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx \\ & = -\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b d}-\frac {2 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d}+\frac {4}{7} \int \frac {1}{\sqrt {d \cos (a+b x)}} \, dx \\ & = -\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b d}-\frac {2 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d}+\frac {\left (4 \sqrt {\cos (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{7 \sqrt {d \cos (a+b x)}} \\ & = \frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b d}-\frac {2 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {d \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b (d \cos (a+b x))^{3/2}} \]

[In]

Integrate[Sin[a + b*x]^4/Sqrt[d*Cos[a + b*x]],x]

[Out]

(d*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/4, 5/2, 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^5)/(5*b*(d*Cos[a + b*x
])^(3/2))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.10

method	result	size
default	\(-\frac {8 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (4 \left (\sin ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-6 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{7 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(208\)

[In]

int(sin(b*x+a)^4/(d*cos(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-8/7*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2*b*x+1/2*a)^8*cos(1/2*b*x+1/2*a)-6*si
n(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a)+(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)))/(-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 = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right ) - 2 i \, \sqrt {2} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, \sqrt {2} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}}{7 \, b d} \]

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/7*(sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 3)*sin(b*x + a) - 2*I*sqrt(2)*sqrt(d)*weierstrassPInverse(-4, 0, c
os(b*x + a) + I*sin(b*x + a)) + 2*I*sqrt(2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)))
/(b*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\text {Timed out} \]

[In]

integrate(sin(b*x+a)**4/(d*cos(b*x+a))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \]

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^4/sqrt(d*cos(b*x + a)), x)

Giac [F]

\[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \]

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^4/sqrt(d*cos(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^4}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]

[In]

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

[Out]

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