\(\int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [145]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 (a+a \sin (c+d x))^{9/2}}{9 a d} \]

[Out]

2/9*(a+a*sin(d*x+c))^(9/2)/a/d

Rubi [A] (verified)

Time = 0.03 (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} \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 (a \sin (c+d x)+a)^{9/2}}{9 a d} \]

[In]

Int[Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(2*(a + a*Sin[c + d*x])^(9/2))/(9*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)^{7/2} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {2 (a+a \sin (c+d x))^{9/2}}{9 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 (a+a \sin (c+d x))^{9/2}}{9 a d} \]

[In]

Integrate[Cos[c + d*x]*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(2*(a + a*Sin[c + d*x])^(9/2))/(9*a*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 a d}\)	\(21\)
default	\(\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 a d}\)	\(21\)

[In]

int(cos(d*x+c)*(a+a*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/9*(a+a*sin(d*x+c))^(9/2)/a/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.08 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} - 4 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{9 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/9*(a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 + 8*a^3 - 4*(a^3*cos(d*x + c)^2 - 2*a^3)*sin(d*x + c))*sqrt(a*s
in(d*x + c) + a)/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (19) = 38\).

Time = 130.57 (sec) , antiderivative size = 156, normalized size of antiderivative = 6.50 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\begin {cases} \frac {2 a^{3} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{4}{\left (c + d x \right )}}{9 d} + \frac {8 a^{3} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{3}{\left (c + d x \right )}}{9 d} + \frac {4 a^{3} \sqrt {a \sin {\left (c + d x \right )} + a} \sin ^{2}{\left (c + d x \right )}}{3 d} + \frac {8 a^{3} \sqrt {a \sin {\left (c + d x \right )} + a} \sin {\left (c + d x \right )}}{9 d} + \frac {2 a^{3} \sqrt {a \sin {\left (c + d x \right )} + a}}{9 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{\frac {7}{2}} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Piecewise((2*a**3*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**4/(9*d) + 8*a**3*sqrt(a*sin(c + d*x) + a)*sin(c + d*x
)**3/(9*d) + 4*a**3*sqrt(a*sin(c + d*x) + a)*sin(c + d*x)**2/(3*d) + 8*a**3*sqrt(a*sin(c + d*x) + a)*sin(c + d
*x)/(9*d) + 2*a**3*sqrt(a*sin(c + d*x) + a)/(9*d), Ne(d, 0)), (x*(a*sin(c) + a)**(7/2)*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}}}{9 \, a d} \]

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/9*(a*sin(d*x + c) + a)^(9/2)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {32 \, \sqrt {2} a^{\frac {7}{2}} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{9 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

32/9*sqrt(2)*a^(7/2)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^9*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [B] (verification not implemented)

Time = 4.46 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^{9/2}}{9\,a\,d} \]

[In]

int(cos(c + d*x)*(a + a*sin(c + d*x))^(7/2),x)

[Out]

(2*(a*(sin(c + d*x) + 1))^(9/2))/(9*a*d)