\(\int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx\) [197]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\frac {4 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{21 b \sqrt {d \cos (a+b x)}}+\frac {4 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{21 b}-\frac {2 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2648, 2715, 2721, 2720} \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\frac {4 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{21 b \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) (d \cos (a+b x))^{5/2}}{7 b d}+\frac {4 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{21 b} \]

[In]

Int[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^2,x]

[Out]

(4*d^2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(21*b*Sqrt[d*Cos[a + b*x]]) + (4*d*Sqrt[d*Cos[a + b*x]]*S
in[a + b*x])/(21*b) - (2*(d*Cos[a + b*x])^(5/2)*Sin[a + b*x])/(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 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))^{5/2} \sin (a+b x)}{7 b d}+\frac {2}{7} \int (d \cos (a+b x))^{3/2} \, dx \\ & = \frac {4 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{21 b}-\frac {2 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b d}+\frac {1}{21} \left (2 d^2\right ) \int \frac {1}{\sqrt {d \cos (a+b x)}} \, dx \\ & = \frac {4 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{21 b}-\frac {2 (d \cos (a+b x))^{5/2} \sin (a+b x)}{7 b d}+\frac {\left (2 d^2 \sqrt {\cos (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{21 \sqrt {d \cos (a+b x)}} \\ & = \frac {4 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{21 b \sqrt {d \cos (a+b x)}}+\frac {4 d \sqrt {d \cos (a+b x)} \sin (a+b x)}{21 b}-\frac {2 (d \cos (a+b x))^{5/2} \sin (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 = 57, normalized size of antiderivative = 0.58 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\frac {(d \cos (a+b x))^{3/2} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \tan ^3(a+b x)}{3 b} \]

[In]

Integrate[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {4 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{2} \left (24 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-60 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+50 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-15 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+\cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{21 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\)	\(208\)

[In]

int((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

4/21*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^2*(24*cos(1/2*b*x+1/2*a)^9-60*cos(1/2*b*x+1/2
*a)^7+50*cos(1/2*b*x+1/2*a)^5-15*cos(1/2*b*x+1/2*a)^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)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))+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 = 90, normalized size of antiderivative = 0.92 \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (i \, \sqrt {2} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - i \, \sqrt {2} d^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (3 \, d \cos \left (b x + a\right )^{2} - 2 \, d\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{21 \, b} \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-2/21*(I*sqrt(2)*d^(3/2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) - I*sqrt(2)*d^(3/2)*weierst
rassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + (3*d*cos(b*x + a)^2 - 2*d)*sqrt(d*cos(b*x + a))*sin(b*x +
a))/b

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\text {Timed out} \]

[In]

integrate((d*cos(b*x+a))**(3/2)*sin(b*x+a)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^2, x)

Giac [F]

\[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{3/2} \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2} \,d x \]

[In]

int(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2),x)

[Out]

int(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2), x)