\(\int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [495]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{9/2}}{9 b^3 d}-\frac {2 (a+b \sin (c+d x))^{11/2}}{11 b^3 d} \]

[Out]

-2/7*(a^2-b^2)*(a+b*sin(d*x+c))^(7/2)/b^3/d+4/9*a*(a+b*sin(d*x+c))^(9/2)/b^3/d-2/11*(a+b*sin(d*x+c))^(11/2)/b^
3/d

Rubi [A] (verified)

Time = 0.12 (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} \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^3 d}-\frac {2 (a+b \sin (c+d x))^{11/2}}{11 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{9/2}}{9 b^3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(7/2))/(7*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(9/2))/(9*b^3*d) - (2*(a + b
*Sin[c + d*x])^(11/2))/(11*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 (a+x)^{5/2} \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 ) (a+x)^{5/2}+2 a (a+x)^{7/2}-(a+x)^{9/2}\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))^{7/2}}{7 b^3 d}+\frac {4 a (a+b \sin (c+d x))^{9/2}}{9 b^3 d}-\frac {2 (a+b \sin (c+d x))^{11/2}}{11 b^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 (a+b \sin (c+d x))^{7/2} \left (8 a^2-99 b^2-28 a b \sin (c+d x)+63 b^2 \sin ^2(c+d x)\right )}{693 b^3 d} \]

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(a + b*Sin[c + d*x])^(7/2)*(8*a^2 - 99*b^2 - 28*a*b*Sin[c + d*x] + 63*b^2*Sin[c + d*x]^2))/(693*b^3*d)

Maple [A] (verified)

Time = 6.90 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {2 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}\right )}{d \,b^{3}}\)	\(62\)
default	\(-\frac {2 \left (\frac {\left (a +b \sin \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {2 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}+\frac {\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}\right )}{d \,b^{3}}\)	\(62\)

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/d/b^3*(1/11*(a+b*sin(d*x+c))^(11/2)-2/9*a*(a+b*sin(d*x+c))^(9/2)+1/7*(a^2-b^2)*(a+b*sin(d*x+c))^(7/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (71) = 142\).

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.72 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (161 \, a b^{4} \cos \left (d x + c\right )^{4} + 8 \, a^{5} - 96 \, a^{3} b^{2} - 136 \, a b^{4} - {\left (3 \, a^{3} b^{2} + 25 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (63 \, b^{5} \cos \left (d x + c\right )^{4} - 4 \, a^{4} b - 184 \, a^{2} b^{3} - 36 \, b^{5} - {\left (113 \, a^{2} b^{3} + 27 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{693 \, b^{3} d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/693*(161*a*b^4*cos(d*x + c)^4 + 8*a^5 - 96*a^3*b^2 - 136*a*b^4 - (3*a^3*b^2 + 25*a*b^4)*cos(d*x + c)^2 + (6
3*b^5*cos(d*x + c)^4 - 4*a^4*b - 184*a^2*b^3 - 36*b^5 - (113*a^2*b^3 + 27*b^5)*cos(d*x + c)^2)*sin(d*x + c))*s
qrt(b*sin(d*x + c) + a)/(b^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (76) = 152\).

Time = 39.22 (sec) , antiderivative size = 391, normalized size of antiderivative = 4.71 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\begin {cases} a^{\frac {5}{2}} x \cos ^{3}{\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\a^{\frac {5}{2}} \cdot \left (\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}\right ) & \text {for}\: b = 0 \\x \left (a + b \sin {\left (c \right )}\right )^{\frac {5}{2}} \cos ^{3}{\left (c \right )} & \text {for}\: d = 0 \\- \frac {16 a^{5} \sqrt {a + b \sin {\left (c + d x \right )}}}{693 b^{3} d} + \frac {8 a^{4} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{693 b^{2} d} + \frac {64 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{231 b d} + \frac {2 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}}{7 b d} + \frac {368 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )}}{693 d} + \frac {6 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac {272 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{4}{\left (c + d x \right )}}{693 d} + \frac {6 a b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} + \frac {8 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{5}{\left (c + d x \right )}}{77 d} + \frac {2 b^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{7 d} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Piecewise((a**(5/2)*x*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (a**(5/2)*(2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c
+ d*x)**2/d), Eq(b, 0)), (x*(a + b*sin(c))**(5/2)*cos(c)**3, Eq(d, 0)), (-16*a**5*sqrt(a + b*sin(c + d*x))/(6
93*b**3*d) + 8*a**4*sqrt(a + b*sin(c + d*x))*sin(c + d*x)/(693*b**2*d) + 64*a**3*sqrt(a + b*sin(c + d*x))*sin(
c + d*x)**2/(231*b*d) + 2*a**3*sqrt(a + b*sin(c + d*x))*cos(c + d*x)**2/(7*b*d) + 368*a**2*sqrt(a + b*sin(c +
d*x))*sin(c + d*x)**3/(693*d) + 6*a**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)*cos(c + d*x)**2/(7*d) + 272*a*b*s
qrt(a + b*sin(c + d*x))*sin(c + d*x)**4/(693*d) + 6*a*b*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2*cos(c + d*x)*
*2/(7*d) + 8*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**5/(77*d) + 2*b**2*sqrt(a + b*sin(c + d*x))*sin(c + d*
x)**3*cos(c + d*x)**2/(7*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (63 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 154 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 99 \, {\left (a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}\right )}}{693 \, b^{3} d} \]

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/693*(63*(b*sin(d*x + c) + a)^(11/2) - 154*(b*sin(d*x + c) + a)^(9/2)*a + 99*(a^2 - b^2)*(b*sin(d*x + c) + a
)^(7/2))/(b^3*d)

Giac [F]

\[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cos(d*x+c)^3*(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)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

int(cos(c + d*x)^3*(a + b*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^3*(a + b*sin(c + d*x))^(5/2), x)