\(\int \sin ^{\frac {7}{2}}(a+b x) \, dx\) [17]

Optimal result

Integrand size = 10, antiderivative size = 70 \[ \int \sin ^{\frac {7}{2}}(a+b x) \, dx=\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{21 b}-\frac {10 \cos (a+b x) \sqrt {\sin (a+b x)}}{21 b}-\frac {2 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{7 b} \]

[Out]

-10/21*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticF(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/
2))/b-2/7*cos(b*x+a)*sin(b*x+a)^(5/2)/b-10/21*cos(b*x+a)*sin(b*x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2715, 2720} \[ \int \sin ^{\frac {7}{2}}(a+b x) \, dx=\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{21 b}-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}-\frac {10 \sqrt {\sin (a+b x)} \cos (a+b x)}{21 b} \]

[In]

Int[Sin[a + b*x]^(7/2),x]

[Out]

(10*EllipticF[(a - Pi/2 + b*x)/2, 2])/(21*b) - (10*Cos[a + b*x]*Sqrt[Sin[a + b*x]])/(21*b) - (2*Cos[a + b*x]*S
in[a + b*x]^(5/2))/(7*b)

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{7 b}+\frac {5}{7} \int \sin ^{\frac {3}{2}}(a+b x) \, dx \\ & = -\frac {10 \cos (a+b x) \sqrt {\sin (a+b x)}}{21 b}-\frac {2 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{7 b}+\frac {5}{21} \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx \\ & = \frac {10 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{21 b}-\frac {10 \cos (a+b x) \sqrt {\sin (a+b x)}}{21 b}-\frac {2 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \sin ^{\frac {7}{2}}(a+b x) \, dx=\frac {-20 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right )+(-23 \cos (a+b x)+3 \cos (3 (a+b x))) \sqrt {\sin (a+b x)}}{42 b} \]

[In]

Integrate[Sin[a + b*x]^(7/2),x]

[Out]

(-20*EllipticF[(-2*a + Pi - 2*b*x)/4, 2] + (-23*Cos[a + b*x] + 3*Cos[3*(a + b*x)])*Sqrt[Sin[a + b*x]])/(42*b)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\frac {2 \left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{7}+\frac {5 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, F\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{21}-\frac {16 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{21}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\)	\(104\)

[In]

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

[Out]

(2/7*cos(b*x+a)^4*sin(b*x+a)+5/21*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticF((
sin(b*x+a)+1)^(1/2),1/2*2^(1/2))-16/21*cos(b*x+a)^2*sin(b*x+a))/cos(b*x+a)/sin(b*x+a)^(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 = 1.24 \[ \int \sin ^{\frac {7}{2}}(a+b x) \, dx=\frac {5 \, \sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 \, \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{3} - 8 \, \cos \left (b x + a\right )\right )} \sqrt {\sin \left (b x + a\right )}}{21 \, b} \]

[In]

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

[Out]

1/21*(5*sqrt(2)*sqrt(-I)*weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a)) + 5*sqrt(2)*sqrt(I)*weierstr
assPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*(3*cos(b*x + a)^3 - 8*cos(b*x + a))*sqrt(sin(b*x + a)))/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate(sin(b*x + a)^(7/2), x)

Giac [F]

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

[In]

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

[Out]

integrate(sin(b*x + a)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.60 \[ \int \sin ^{\frac {7}{2}}(a+b x) \, dx=-\frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{9/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{9/4}} \]

[In]

int(sin(a + b*x)^(7/2),x)

[Out]

-(cos(a + b*x)*sin(a + b*x)^(9/2)*hypergeom([-5/4, 1/2], 3/2, cos(a + b*x)^2))/(b*(sin(a + b*x)^2)^(9/4))