\(\int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [517]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 \left (a^2-b^2\right )}{b^3 d \sqrt {a+b \sin (c+d x)}}+\frac {4 a \sqrt {a+b \sin (c+d x)}}{b^3 d}-\frac {2 (a+b \sin (c+d x))^{3/2}}{3 b^3 d} \]

[Out]

-2/3*(a+b*sin(d*x+c))^(3/2)/b^3/d+2*(a^2-b^2)/b^3/d/(a+b*sin(d*x+c))^(1/2)+4*a*(a+b*sin(d*x+c))^(1/2)/b^3/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, 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} \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 \left (a^2-b^2\right )}{b^3 d \sqrt {a+b \sin (c+d x)}}-\frac {2 (a+b \sin (c+d x))^{3/2}}{3 b^3 d}+\frac {4 a \sqrt {a+b \sin (c+d x)}}{b^3 d} \]

[In]

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

[Out]

(2*(a^2 - b^2))/(b^3*d*Sqrt[a + b*Sin[c + d*x]]) + (4*a*Sqrt[a + b*Sin[c + d*x]])/(b^3*d) - (2*(a + b*Sin[c +
d*x])^(3/2))/(3*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 \frac {b^2-x^2}{(a+x)^{3/2}} \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {-a^2+b^2}{(a+x)^{3/2}}+\frac {2 a}{\sqrt {a+x}}-\sqrt {a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {2 \left (a^2-b^2\right )}{b^3 d \sqrt {a+b \sin (c+d x)}}+\frac {4 a \sqrt {a+b \sin (c+d x)}}{b^3 d}-\frac {2 (a+b \sin (c+d x))^{3/2}}{3 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {16 a^2-7 b^2+b^2 \cos (2 (c+d x))+8 a b \sin (c+d x)}{3 b^3 d \sqrt {a+b \sin (c+d x)}} \]

[In]

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

[Out]

(16*a^2 - 7*b^2 + b^2*Cos[2*(c + d*x)] + 8*a*b*Sin[c + d*x])/(3*b^3*d*Sqrt[a + b*Sin[c + d*x]])

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (b^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + 8 \, a^{2} - 4 \, b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{3 \, {\left (b^{4} d \sin \left (d x + c\right ) + a b^{3} d\right )}} \]

[In]

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

[Out]

2/3*(b^2*cos(d*x + c)^2 + 4*a*b*sin(d*x + c) + 8*a^2 - 4*b^2)*sqrt(b*sin(d*x + c) + a)/(b^4*d*sin(d*x + c) + a
*b^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a}{b^{2}} - \frac {3 \, {\left (a^{2} - b^{2}\right )}}{\sqrt {b \sin \left (d x + c\right ) + a} b^{2}}\right )}}{3 \, b d} \]

[In]

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

[Out]

-2/3*(((b*sin(d*x + c) + a)^(3/2) - 6*sqrt(b*sin(d*x + c) + a)*a)/b^2 - 3*(a^2 - b^2)/(sqrt(b*sin(d*x + c) + a
)*b^2))/(b*d)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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