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]

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

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Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.18, size = 58, normalized size = 0.70 \[ \frac {(a+b \sin (c+d x))^{5/2} \left (-16 a^2+40 a b \sin (c+d x)+35 b^2 \cos (2 (c+d x))+91 b^2\right )}{315 b^3 d} \]

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)

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fricas [A]  time = 1.01, size = 111, normalized size = 1.34 \[ -\frac {2 \, {\left (35 \, b^{4} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 60 \, a^{2} b^{2} - 28 \, b^{4} - {\left (3 \, a^{2} b^{2} + 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (25 \, a b^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} b + 38 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{315 \, b^{3} d} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{3}\,{d x} \]

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)

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maple [A]  time = 0.33, size = 55, normalized size = 0.66 \[ -\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}} \left (-35 b^{2} \left (\cos ^{2}\left (d x +c \right )\right )-20 a b \sin \left (d x +c \right )+8 a^{2}-28 b^{2}\right )}{315 b^{3} d} \]

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

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maxima [A]  time = 0.32, size = 61, normalized size = 0.73 \[ -\frac {2 \, {\left (35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 90 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 63 \, {\left (a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\right )}}{315 \, b^{3} d} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 121.57, size = 314, normalized size = 3.78 \[ \begin {cases} a^{\frac {3}{2}} x \cos ^{3}{\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\a^{\frac {3}{2}} \left (\frac {2 \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d}\right ) & \text {for}\: b = 0 \\x \left (a + b \sin {\relax (c )}\right )^{\frac {3}{2}} \cos ^{3}{\relax (c )} & \text {for}\: d = 0 \\- \frac {16 a^{4} \sqrt {a + b \sin {\left (c + d x \right )}}}{315 b^{3} d} + \frac {8 a^{3} \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{315 b^{2} d} + \frac {8 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{21 b d} + \frac {2 a^{2} \sqrt {a + b \sin {\left (c + d x \right )}} \cos ^{2}{\left (c + d x \right )}}{5 b d} + \frac {152 a \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{3}{\left (c + d x \right )}}{315 d} + \frac {4 a \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{4}{\left (c + d x \right )}}{45 d} + \frac {2 b \sqrt {a + b \sin {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} & \text {otherwise} \end {cases} \]

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)
)

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