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

Optimal. Leaf size=83 \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^3 d}-\frac{2 (a+b \sin (c+d x))^{11/2}}{11 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{9/2}}{9 b^3 d} \]

[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)

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Rubi [A]  time = 0.0910357, 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} \[ -\frac{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^3 d}-\frac{2 (a+b \sin (c+d x))^{11/2}}{11 b^3 d}+\frac{4 a (a+b \sin (c+d x))^{9/2}}{9 b^3 d} \]

Antiderivative was successfully verified.

[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 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))^{5/2} \, dx &=\frac{\operatorname{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{\operatorname{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]  time = 0.0858061, size = 58, normalized size = 0.7 \[ -\frac{2 (a+b \sin (c+d x))^{7/2} \left (8 a^2-28 a b \sin (c+d x)+63 b^2 \sin ^2(c+d x)-99 b^2\right )}{693 b^3 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.224, size = 55, normalized size = 0.7 \begin{align*} -{\frac{-126\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-56\,ab\sin \left ( dx+c \right ) +16\,{a}^{2}-72\,{b}^{2}}{693\,{b}^{3}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{7}{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))^(5/2),x)

[Out]

-2/693/b^3*(a+b*sin(d*x+c))^(7/2)*(-63*b^2*cos(d*x+c)^2-28*a*b*sin(d*x+c)+8*a^2-36*b^2)/d

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Maxima [A]  time = 0.951021, size = 82, normalized size = 0.99 \begin{align*} -\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} \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))^(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)

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Fricas [B]  time = 2.25444, size = 342, normalized size = 4.12 \begin{align*} -\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} \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))^(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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="giac")

[Out]

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