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

Optimal. Leaf size=83 \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^3 d}-\frac{2 (a+b \sin (c+d x))^{9/2}}{9 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{7/2}}{7 b^3 d} \]

[Out]

(-2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(5/2))/(5*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(7/2))/(7*b^3*d) - (2*(a + b
*Sin[c + d*x])^(9/2))/(9*b^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0913597, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2668, 697} \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^3 d}-\frac{2 (a+b \sin (c+d x))^{9/2}}{9 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{7/2}}{7 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a^2 - b^2)*(a + b*Sin[c + d*x])^(5/2))/(5*b^3*d) + (4*a*(a + b*Sin[c + d*x])^(7/2))/(7*b^3*d) - (2*(a + b
*Sin[c + d*x])^(9/2))/(9*b^3*d)

Rule 2668

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]

Rule 697

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]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^{3/2} \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (-a^2+b^2\right ) (a+x)^{3/2}+2 a (a+x)^{5/2}-(a+x)^{7/2}\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))^{5/2}}{5 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{7/2}}{7 b^3 d}-\frac{2 (a+b \sin (c+d x))^{9/2}}{9 b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.181559, size = 58, normalized size = 0.7 \[ \frac{(a+b \sin (c+d x))^{5/2} \left (-16 a^2+40 a b \sin (c+d x)+35 b^2 \cos (2 (c+d x))+91 b^2\right )}{315 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sin[c + d*x])^(5/2)*(-16*a^2 + 91*b^2 + 35*b^2*Cos[2*(c + d*x)] + 40*a*b*Sin[c + d*x]))/(315*b^3*d)

________________________________________________________________________________________

Maple [A]  time = 0.22, size = 55, normalized size = 0.7 \begin{align*} -{\frac{-70\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-40\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}-56\,{b}^{2}}{315\,{b}^{3}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/b^3*(a+b*sin(d*x+c))^(5/2)*(-35*b^2*cos(d*x+c)^2-20*a*b*sin(d*x+c)+8*a^2-28*b^2)/d

________________________________________________________________________________________

Maxima [A]  time = 0.961138, size = 82, normalized size = 0.99 \begin{align*} -\frac{2 \,{\left (35 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} - 90 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a + 63 \,{\left (a^{2} - b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\right )}}{315 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*(b*sin(d*x + c) + a)^(9/2) - 90*(b*sin(d*x + c) + a)^(7/2)*a + 63*(a^2 - b^2)*(b*sin(d*x + c) + a)^
(5/2))/(b^3*d)

________________________________________________________________________________________

Fricas [A]  time = 2.49276, size = 265, normalized size = 3.19 \begin{align*} -\frac{2 \,{\left (35 \, b^{4} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 60 \, a^{2} b^{2} - 28 \, b^{4} -{\left (3 \, a^{2} b^{2} + 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (25 \, a b^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} b + 38 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}}{315 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*b^4*cos(d*x + c)^4 + 8*a^4 - 60*a^2*b^2 - 28*b^4 - (3*a^2*b^2 + 7*b^4)*cos(d*x + c)^2 - 2*(25*a*b^3
*cos(d*x + c)^2 + 2*a^3*b + 38*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^3*d)

________________________________________________________________________________________

Sympy [A]  time = 134.232, size = 314, normalized size = 3.78 \begin{align*} \begin{cases} a^{\frac{3}{2}} x \cos ^{3}{\left (c \right )} & \text{for}\: b = 0 \wedge d = 0 \\a^{\frac{3}{2}} \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{3}{2}} \cos ^{3}{\left (c \right )} & \text{for}\: d = 0 \\- \frac{16 a^{4} \sqrt{a + b \sin{\left (c + d x \right )}}}{315 b^{3} d} + \frac{8 a^{3} \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )}}{315 b^{2} d} + \frac{8 a^{2} \sqrt{a + b \sin{\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{21 b d} + \frac{2 a^{2} \sqrt{a + b \sin{\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}}{5 b d} + \frac{152 a \sqrt{a + b \sin{\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )}}{315 d} + \frac{4 a \sqrt{a + b \sin{\left (c + d x \right )}} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac{8 b \sqrt{a + b \sin{\left (c + d x \right )}} \sin ^{4}{\left (c + d x \right )}}{45 d} + \frac{2 b \sqrt{a + b \sin{\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**(3/2)*x*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (a**(3/2)*(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))**(3/2)*cos(c)**3, Eq(d, 0)), (-16*a**4*sqrt(a + b*sin(c + d*x))/(3
15*b**3*d) + 8*a**3*sqrt(a + b*sin(c + d*x))*sin(c + d*x)/(315*b**2*d) + 8*a**2*sqrt(a + b*sin(c + d*x))*sin(c
+ d*x)**2/(21*b*d) + 2*a**2*sqrt(a + b*sin(c + d*x))*cos(c + d*x)**2/(5*b*d) + 152*a*sqrt(a + b*sin(c + d*x))
*sin(c + d*x)**3/(315*d) + 4*a*sqrt(a + b*sin(c + d*x))*sin(c + d*x)*cos(c + d*x)**2/(5*d) + 8*b*sqrt(a + b*si
n(c + d*x))*sin(c + d*x)**4/(45*d) + 2*b*sqrt(a + b*sin(c + d*x))*sin(c + d*x)**2*cos(c + d*x)**2/(5*d), True)
)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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