3.3.20 \(\int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx\) [220]

Optimal. Leaf size=102 \[ \frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{7 b d^4 \sqrt {d \cos (a+b x)}}-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2646, 2721, 2720} \begin {gather*} \frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{7 b d^4 \sqrt {d \cos (a+b x)}}-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^4/(d*Cos[a + b*x])^(9/2),x]

[Out]

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

Rule 2646

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

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*} \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx &=\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}} \, dx}{7 d^2}\\ &=-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}+\frac {4 \int \frac {1}{\sqrt {d \cos (a+b x)}} \, dx}{7 d^4}\\ &=-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}+\frac {\left (4 \sqrt {\cos (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{7 d^4 \sqrt {d \cos (a+b x)}}\\ &=\frac {8 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{7 b d^4 \sqrt {d \cos (a+b x)}}-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 65, normalized size = 0.64 \begin {gather*} \frac {\cos ^3(a+b x) \cos ^2(a+b x)^{3/4} \, _2F_1\left (\frac {5}{2},\frac {11}{4};\frac {7}{2};\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b (d \cos (a+b x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^4/(d*Cos[a + b*x])^(9/2),x]

[Out]

(Cos[a + b*x]^3*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[5/2, 11/4, 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^5)/(5*b*
(d*Cos[a + b*x])^(9/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(114)=228\).
time = 0.29, size = 398, normalized size = 3.90

method	result	size
default	\(\frac {8 \left (8 \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 ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \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 ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\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 )+6 \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 )+6 \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 ) \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 )}}{7 d^{4} \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\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}\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^4/(d*cos(b*x+a))^(9/2),x,method=_RETURNVERBOSE)

[Out]

8/7*(8*(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)^6-12*(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)^4-6*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+6*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*si
n(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+6*cos(1/2*b*x+1/2*a)*si
n(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^4*(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)^2-1)^3/(-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

________________________________________________________________________________________

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(sin(b*x+a)^4/(d*cos(b*x+a))^(9/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(9/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 114, normalized size = 1.12 \begin {gather*} -\frac {2 \, {\left (2 i \, \sqrt {2} \sqrt {d} \cos \left (b x + a\right )^{4} {\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} \cos \left (b x + a\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + \sqrt {d \cos \left (b x + a\right )} {\left (3 \, \cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )}}{7 \, b d^{5} \cos \left (b x + a\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4/(d*cos(b*x+a))^(9/2),x, algorithm="fricas")

[Out]

-2/7*(2*I*sqrt(2)*sqrt(d)*cos(b*x + a)^4*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) - 2*I*sqrt(
2)*sqrt(d)*cos(b*x + a)^4*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + sqrt(d*cos(b*x + a))*(3*
cos(b*x + a)^2 - 1)*sin(b*x + a))/(b*d^5*cos(b*x + a)^4)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**4/(d*cos(b*x+a))**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4847 deep

________________________________________________________________________________________

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(sin(b*x+a)^4/(d*cos(b*x+a))^(9/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^4/(d*cos(b*x + a))^(9/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/(d*cos(a + b*x))^(9/2),x)

[Out]

int(sin(a + b*x)^4/(d*cos(a + b*x))^(9/2), x)

________________________________________________________________________________________