\(\int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx\) [213]

Optimal result

Integrand size = 21, antiderivative size = 128 \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\frac {8 d^2 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{65 b \sqrt {\cos (a+b x)}}+\frac {8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac {4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac {2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d} \]

[Out]

8/195*d*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b-4/39*(d*cos(b*x+a))^(7/2)*sin(b*x+a)/b/d-2/13*(d*cos(b*x+a))^(7/2)*s
in(b*x+a)^3/b/d+8/65*d^2*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))
*(d*cos(b*x+a))^(1/2)/b/cos(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 128, 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, 2719} \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\frac {8 d^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{65 b \sqrt {\cos (a+b x)}}-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{7/2}}{13 b d}-\frac {4 \sin (a+b x) (d \cos (a+b x))^{7/2}}{39 b d}+\frac {8 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{195 b} \]

[In]

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

[Out]

(8*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(65*b*Sqrt[Cos[a + b*x]]) + (8*d*(d*Cos[a + b*x])^(3/2)
*Sin[a + b*x])/(195*b) - (4*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(39*b*d) - (2*(d*Cos[a + b*x])^(7/2)*Sin[a +
b*x]^3)/(13*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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac {6}{13} \int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx \\ & = -\frac {4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac {2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac {4}{39} \int (d \cos (a+b x))^{5/2} \, dx \\ & = \frac {8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac {4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac {2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac {1}{65} \left (4 d^2\right ) \int \sqrt {d \cos (a+b x)} \, dx \\ & = \frac {8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac {4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac {2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 b d}+\frac {\left (4 d^2 \sqrt {d \cos (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \, dx}{65 \sqrt {\cos (a+b x)}} \\ & = \frac {8 d^2 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{65 b \sqrt {\cos (a+b x)}}+\frac {8 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{195 b}-\frac {4 (d \cos (a+b x))^{7/2} \sin (a+b x)}{39 b d}-\frac {2 (d \cos (a+b x))^{7/2} \sin ^3(a+b x)}{13 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 = 65, normalized size of antiderivative = 0.51 \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\frac {(d \cos (a+b x))^{5/2} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^2(a+b x) \tan ^3(a+b x)}{5 b} \]

[In]

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

[Out]

((d*Cos[a + b*x])^(5/2)*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, 5/2, 7/2, Sin[a + b*x]^2]*Sin[a + b*x]^
2*Tan[a + b*x]^3)/(5*b)

Maple [A] (verified)

Time = 11.09 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.95

method	result	size
default	\(-\frac {8 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{3} \left (480 \left (\cos ^{15}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1920 \left (\cos ^{13}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3040 \left (\cos ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2400 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+958 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-156 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+3 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{195 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(249\)

[In]

int((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

-8/195*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^3*(480*cos(1/2*b*x+1/2*a)^15-1920*cos(1/2*b
*x+1/2*a)^13+3040*cos(1/2*b*x+1/2*a)^11-2400*cos(1/2*b*x+1/2*a)^9+958*cos(1/2*b*x+1/2*a)^7-156*cos(1/2*b*x+1/2
*a)^5-5*cos(1/2*b*x+1/2*a)^3-3*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(1-2*cos(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2
*b*x+1/2*a),2^(1/2))+3*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*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 = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=-\frac {2 \, {\left (-6 i \, \sqrt {2} d^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 6 i \, \sqrt {2} d^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) - {\left (15 \, d^{2} \cos \left (b x + a\right )^{5} - 25 \, d^{2} \cos \left (b x + a\right )^{3} + 4 \, d^{2} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{195 \, b} \]

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-2/195*(-6*I*sqrt(2)*d^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)))
+ 6*I*sqrt(2)*d^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) - (15
*d^2*cos(b*x + a)^5 - 25*d^2*cos(b*x + a)^3 + 4*d^2*cos(b*x + a))*sqrt(d*cos(b*x + a))*sin(b*x + a))/b

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\text {Timed out} \]

[In]

integrate((d*cos(b*x+a))**(5/2)*sin(b*x+a)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{4} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(5/2)*sin(b*x + a)^4, x)

Giac [F]

\[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{4} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(5/2)*sin(b*x+a)^4,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(5/2)*sin(b*x + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{5/2} \sin ^4(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2} \,d x \]

[In]

int(sin(a + b*x)^4*(d*cos(a + b*x))^(5/2),x)

[Out]

int(sin(a + b*x)^4*(d*cos(a + b*x))^(5/2), x)