\(\int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [120]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 (a+a \sin (c+d x))^{5/2}}{5 a d} \]

[Out]

2/5*(a+a*sin(d*x+c))^(5/2)/a/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 32} \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 (a \sin (c+d x)+a)^{5/2}}{5 a d} \]

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^(5/2))/(5*a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2746

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 (a+a \sin (c+d x))^{5/2}}{5 a d} \]

[In]

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

[Out]

(2*(a + a*Sin[c + d*x])^(5/2))/(5*a*d)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 a d}\)	\(21\)
default	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 a d}\)	\(21\)

[In]

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

[Out]

2/5*(a+a*sin(d*x+c))^(5/2)/a/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (a \cos \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - 2 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, d} \]

[In]

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

[Out]

-2/5*(a*cos(d*x + c)^2 - 2*a*sin(d*x + c) - 2*a)*sqrt(a*sin(d*x + c) + a)/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (19) = 38\).

Time = 1.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.75 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\begin {cases} \frac {2 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{5 d} + \frac {4 a \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{5 d} + \frac {2 a \sqrt {a \sin {\left (c + d x \right )} + a}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{\frac {3}{2}} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2/(5*d) + 4*a*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)/(5*d
) + 2*a*sqrt(a*sin(c + d*x) + a)/(5*d), Ne(d, 0)), (x*(a*sin(c) + a)**(3/2)*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{5 \, a d} \]

[In]

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

[Out]

2/5*(a*sin(d*x + c) + a)^(5/2)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {8 \, \sqrt {2} a^{\frac {3}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{5 \, d} \]

[In]

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

[Out]

8/5*sqrt(2)*a^(3/2)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^{5/2}}{5\,a\,d} \]

[In]

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

[Out]

(2*(a*(sin(c + d*x) + 1))^(5/2))/(5*a*d)