3.473 \(\int \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=154 \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac{2 (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{9/2}}{9 b^5 d} \]

[Out]

(2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d
) + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(9/2))/(9*b^5*d) + (2*(
a + b*Sin[c + d*x])^(11/2))/(11*b^5*d)

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Rubi [A]  time = 0.11659, antiderivative size = 154, 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} \[ \frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac{2 (a+b \sin (c+d x))^{11/2}}{11 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{9/2}}{9 b^5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d
) + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(9/2))/(9*b^5*d) + (2*(
a + b*Sin[c + d*x])^(11/2))/(11*b^5*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 ^5(c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+x} \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (a^2-b^2\right )^2 \sqrt{a+x}-4 \left (a^3-a b^2\right ) (a+x)^{3/2}+2 \left (3 a^2-b^2\right ) (a+x)^{5/2}-4 a (a+x)^{7/2}+(a+x)^{9/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac{8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac{4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}-\frac{8 a (a+b \sin (c+d x))^{9/2}}{9 b^5 d}+\frac{2 (a+b \sin (c+d x))^{11/2}}{11 b^5 d}\\ \end{align*}

Mathematica [A]  time = 0.324085, size = 117, normalized size = 0.76 \[ \frac{2 (a+b \sin (c+d x))^{3/2} \left (8 \left (15 b^2 \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)+\left (99 a b^3-24 a^3 b\right ) \sin (c+d x)-66 a^2 b^2+16 a^4-35 a b^3 \sin ^3(c+d x)+105 b^4\right )+315 b^4 \cos ^4(c+d x)\right )}{3465 b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(2*(a + b*Sin[c + d*x])^(3/2)*(315*b^4*Cos[c + d*x]^4 + 8*(16*a^4 - 66*a^2*b^2 + 105*b^4 + (-24*a^3*b + 99*a*b
^3)*Sin[c + d*x] + 15*b^2*(2*a^2 - 3*b^2)*Sin[c + d*x]^2 - 35*a*b^3*Sin[c + d*x]^3)))/(3465*b^5*d)

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Maple [A]  time = 0.346, size = 126, normalized size = 0.8 \begin{align*}{\frac{630\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{4}+560\,a{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -480\,{a}^{2}{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+720\,{b}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-384\,{a}^{3}b\sin \left ( dx+c \right ) +1024\,a{b}^{3}\sin \left ( dx+c \right ) +256\,{a}^{4}-576\,{a}^{2}{b}^{2}+960\,{b}^{4}}{3465\,{b}^{5}d} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^(1/2),x)

[Out]

2/3465/b^5*(a+b*sin(d*x+c))^(3/2)*(315*b^4*cos(d*x+c)^4+280*a*b^3*cos(d*x+c)^2*sin(d*x+c)-240*a^2*b^2*cos(d*x+
c)^2+360*b^4*cos(d*x+c)^2-192*a^3*b*sin(d*x+c)+512*a*b^3*sin(d*x+c)+128*a^4-288*a^2*b^2+480*b^4)/d

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Maxima [A]  time = 0.969357, size = 157, normalized size = 1.02 \begin{align*} \frac{2 \,{\left (315 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a + 990 \,{\left (3 \, a^{2} - b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} - 2772 \,{\left (a^{3} - a b^{2}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} + 1155 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\right )}}{3465 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/3465*(315*(b*sin(d*x + c) + a)^(11/2) - 1540*(b*sin(d*x + c) + a)^(9/2)*a + 990*(3*a^2 - b^2)*(b*sin(d*x + c
) + a)^(7/2) - 2772*(a^3 - a*b^2)*(b*sin(d*x + c) + a)^(5/2) + 1155*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) +
a)^(3/2))/(b^5*d)

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Fricas [A]  time = 3.58679, size = 348, normalized size = 2.26 \begin{align*} \frac{2 \,{\left (35 \, a b^{4} \cos \left (d x + c\right )^{4} + 128 \, a^{5} - 480 \, a^{3} b^{2} + 992 \, a b^{4} - 16 \,{\left (3 \, a^{3} b^{2} - 8 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left (315 \, b^{5} \cos \left (d x + c\right )^{4} - 64 \, a^{4} b + 224 \, a^{2} b^{3} + 480 \, b^{5} + 40 \,{\left (a^{2} b^{3} + 9 \, 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}}{3465 \, b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(35*a*b^4*cos(d*x + c)^4 + 128*a^5 - 480*a^3*b^2 + 992*a*b^4 - 16*(3*a^3*b^2 - 8*a*b^4)*cos(d*x + c)^2
+ (315*b^5*cos(d*x + c)^4 - 64*a^4*b + 224*a^2*b^3 + 480*b^5 + 40*(a^2*b^3 + 9*b^5)*cos(d*x + c)^2)*sin(d*x +
c))*sqrt(b*sin(d*x + c) + a)/(b^5*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)**5*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.45228, size = 234, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (1155 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} + \frac{315 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{11}{2}}}{b^{4}} - \frac{1540 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{9}{2}} a}{b^{4}} + \frac{2970 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}} a^{2}}{b^{4}} - \frac{2772 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{3}}{b^{4}} + \frac{1155 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{4}}{b^{4}} - \frac{990 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}{b^{2}} + \frac{2772 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a}{b^{2}} - \frac{2310 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{2}}{b^{2}}\right )}}{3465 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/3465*(1155*(b*sin(d*x + c) + a)^(3/2) + 315*(b*sin(d*x + c) + a)^(11/2)/b^4 - 1540*(b*sin(d*x + c) + a)^(9/2
)*a/b^4 + 2970*(b*sin(d*x + c) + a)^(7/2)*a^2/b^4 - 2772*(b*sin(d*x + c) + a)^(5/2)*a^3/b^4 + 1155*(b*sin(d*x
+ c) + a)^(3/2)*a^4/b^4 - 990*(b*sin(d*x + c) + a)^(7/2)/b^2 + 2772*(b*sin(d*x + c) + a)^(5/2)*a/b^2 - 2310*(b
*sin(d*x + c) + a)^(3/2)*a^2/b^2)/(b*d)