\(\int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx\) [195]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\frac {20 d^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{231 b \sqrt {d \cos (a+b x)}}+\frac {20 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac {4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac {2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d} \]

[Out]

4/77*d*(d*cos(b*x+a))^(5/2)*sin(b*x+a)/b-2/11*(d*cos(b*x+a))^(9/2)*sin(b*x+a)/b/d+20/231*d^4*(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
0/231*d^3*sin(b*x+a)*(d*cos(b*x+a))^(1/2)/b

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2648, 2715, 2721, 2720} \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\frac {20 d^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{231 b \sqrt {d \cos (a+b x)}}+\frac {20 d^3 \sin (a+b x) \sqrt {d \cos (a+b x)}}{231 b}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{9/2}}{11 b d}+\frac {4 d \sin (a+b x) (d \cos (a+b x))^{5/2}}{77 b} \]

[In]

Int[(d*Cos[a + b*x])^(7/2)*Sin[a + b*x]^2,x]

[Out]

(20*d^4*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(231*b*Sqrt[d*Cos[a + b*x]]) + (20*d^3*Sqrt[d*Cos[a + b*
x]]*Sin[a + b*x])/(231*b) + (4*d*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(77*b) - (2*(d*Cos[a + b*x])^(9/2)*Sin[a
+ b*x])/(11*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 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac {2}{11} \int (d \cos (a+b x))^{7/2} \, dx \\ & = \frac {4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac {2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac {1}{77} \left (10 d^2\right ) \int (d \cos (a+b x))^{3/2} \, dx \\ & = \frac {20 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac {4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac {2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac {1}{231} \left (10 d^4\right ) \int \frac {1}{\sqrt {d \cos (a+b x)}} \, dx \\ & = \frac {20 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac {4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac {2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d}+\frac {\left (10 d^4 \sqrt {\cos (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{231 \sqrt {d \cos (a+b x)}} \\ & = \frac {20 d^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{231 b \sqrt {d \cos (a+b x)}}+\frac {20 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{231 b}+\frac {4 d (d \cos (a+b x))^{5/2} \sin (a+b x)}{77 b}-\frac {2 (d \cos (a+b x))^{9/2} \sin (a+b x)}{11 b d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.48 \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\frac {d^2 (d \cos (a+b x))^{3/2} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \tan ^3(a+b x)}{3 b} \]

[In]

Integrate[(d*Cos[a + b*x])^(7/2)*Sin[a + b*x]^2,x]

[Out]

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

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.87

method	result	size
default	\(\frac {4 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{4} \left (672 \left (\cos ^{13}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2352 \left (\cos ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3312 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2400 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+922 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-159 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+5 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{231 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(236\)

[In]

int((d*cos(b*x+a))^(7/2)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

4/231*(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*(672*cos(1/2*b*x+1/2*a)^13-2352*cos(1/2*b*
x+1/2*a)^11+3312*cos(1/2*b*x+1/2*a)^9-2400*cos(1/2*b*x+1/2*a)^7+922*cos(1/2*b*x+1/2*a)^5-159*cos(1/2*b*x+1/2*a
)^3-5*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(1-2*cos(1/2*b*x+1/2*a)^2)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))+5*co
s(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 = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.85 \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} d^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 5 i \, \sqrt {2} d^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (21 \, d^{3} \cos \left (b x + a\right )^{4} - 6 \, d^{3} \cos \left (b x + a\right )^{2} - 10 \, d^{3}\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{231 \, b} \]

[In]

integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-2/231*(5*I*sqrt(2)*d^(7/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) - 5*I*sqrt(2)*d^(7/2)*we
ierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + (21*d^3*cos(b*x + a)^4 - 6*d^3*cos(b*x + a)^2 - 10*d
^3)*sqrt(d*cos(b*x + a))*sin(b*x + a))/b

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\text {Timed out} \]

[In]

integrate((d*cos(b*x+a))**(7/2)*sin(b*x+a)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(7/2)*sin(b*x + a)^2, x)

Giac [F]

\[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(7/2)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(7/2)*sin(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{7/2} \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2} \,d x \]

[In]

int(sin(a + b*x)^2*(d*cos(a + b*x))^(7/2),x)

[Out]

int(sin(a + b*x)^2*(d*cos(a + b*x))^(7/2), x)