\(\int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) [474]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d}-\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d} \]

[Out]

-2/3*(a^2-b^2)*(a+b*sin(d*x+c))^(3/2)/b^3/d+4/5*a*(a+b*sin(d*x+c))^(5/2)/b^3/d-2/7*(a+b*sin(d*x+c))^(7/2)/b^3/
d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2747, 711} \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}-\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d} \]

[In]

Int[Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(5/2))/(5*b^3*d) - (2*(a + b
*Sin[c + d*x])^(7/2))/(7*b^3*d)

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \sqrt {a+x} \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \left (\left (-a^2+b^2\right ) \sqrt {a+x}+2 a (a+x)^{3/2}-(a+x)^{5/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = -\frac {2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{5/2}}{5 b^3 d}-\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {(a+b \sin (c+d x))^{3/2} \left (-16 a^2+55 b^2+15 b^2 \cos (2 (c+d x))+24 a b \sin (c+d x)\right )}{105 b^3 d} \]

[In]

Integrate[Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

((a + b*Sin[c + d*x])^(3/2)*(-16*a^2 + 55*b^2 + 15*b^2*Cos[2*(c + d*x)] + 24*a*b*Sin[c + d*x]))/(105*b^3*d)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,b^{3}}\)	\(62\)
default	\(-\frac {2 \left (\frac {\left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,b^{3}}\)	\(62\)

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/d/b^3*(1/7*(a+b*sin(d*x+c))^(7/2)-2/5*a*(a+b*sin(d*x+c))^(5/2)+1/3*(a^2-b^2)*(a+b*sin(d*x+c))^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - 8 \, a^{3} + 32 \, a b^{2} + {\left (15 \, b^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{2} b + 20 \, b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{105 \, b^{3} d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/105*(3*a*b^2*cos(d*x + c)^2 - 8*a^3 + 32*a*b^2 + (15*b^3*cos(d*x + c)^2 + 4*a^2*b + 20*b^3)*sin(d*x + c))*sq
rt(b*sin(d*x + c) + a)/(b^3*d)

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**3*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (15 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 42 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 35 \, {\left (a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}\right )}}{105 \, b^{3} d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/105*(15*(b*sin(d*x + c) + a)^(7/2) - 42*(b*sin(d*x + c) + a)^(5/2)*a + 35*(a^2 - b^2)*(b*sin(d*x + c) + a)^
(3/2))/(b^3*d)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (15 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 42 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{2}\right )}}{105 \, b^{3} d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/105*(15*(b*sin(d*x + c) + a)^(7/2) - 42*(b*sin(d*x + c) + a)^(5/2)*a + 35*(b*sin(d*x + c) + a)^(3/2)*a^2 -
35*(b*sin(d*x + c) + a)^(3/2)*b^2)/(b^3*d)

Mupad [F(-1)]

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

[In]

int(cos(c + d*x)^3*(a + b*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^3*(a + b*sin(c + d*x))^(1/2), x)