\(\int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [496]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

[Out]

2/7*(a+b*sin(d*x+c))^(7/2)/b/d

Rubi [A] (verified)

Time = 0.05 (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 = {2747, 32} \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

[In]

Int[Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(2*(a + b*Sin[c + d*x])^(7/2))/(7*b*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 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 (a+x)^{5/2} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = \frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b d} \]

[In]

Integrate[Cos[c + d*x]*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 b d}\)	\(21\)
default	\(\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 b d}\)	\(21\)

[In]

int(cos(d*x+c)*(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/7*(a+b*sin(d*x+c))^(7/2)/b/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (20) = 40\).

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 3.21 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{7 \, b d} \]

[In]

integrate(cos(d*x+c)*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (19) = 38\).

Time = 13.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 6.25 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\begin {cases} a^{\frac {5}{2}} x \cos {\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {a^{\frac {5}{2}} \sin {\left (c + d x \right )}}{d} & \text {for}\: b = 0 \\x \left (a + b \sin {\left (c \right )}\right )^{\frac {5}{2}} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {2 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}}}{7 b d} + \frac {6 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{7 d} + \frac {6 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{7 d} + \frac {2 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )}}{7 d} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Piecewise((a**(5/2)*x*cos(c), Eq(b, 0) & Eq(d, 0)), (a**(5/2)*sin(c + d*x)/d, Eq(b, 0)), (x*(a + b*sin(c))**(5
/2)*cos(c), Eq(d, 0)), (2*a**3*sqrt(a + b*sin(c + d*x))/(7*b*d) + 6*a**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)
/(7*d) + 6*a*b*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2/(7*d) + 2*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**
3/(7*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}{7 \, b d} \]

[In]

integrate(cos(d*x+c)*(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/7*(b*sin(d*x + c) + a)^(7/2)/(b*d)

Giac [F]

\[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)*(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^(5/2)*cos(d*x + c), x)

Mupad [B] (verification not implemented)

Time = 5.36 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\frac {2\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{7/2}}{7\,b\,d} \]

[In]

int(cos(c + d*x)*(a + b*sin(c + d*x))^(5/2),x)

[Out]

(2*(a + b*sin(c + d*x))^(7/2))/(7*b*d)