3.5.73 \(\int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx\) [473]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 154, 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 = {2747, 711} \begin {gather*} \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} \end {gather*}

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 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*} \int \cos ^5(c+d x) \sqrt {a+b \sin (c+d x)} \, dx &=\frac {\text {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 {\text {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*}

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Mathematica [A]
time = 0.33, size = 117, normalized size = 0.76 \begin {gather*} \frac {2 (a+b \sin (c+d x))^{3/2} \left (315 b^4 \cos ^4(c+d x)+8 \left (16 a^4-66 a^2 b^2+105 b^4+\left (-24 a^3 b+99 a b^3\right ) \sin (c+d x)+15 b^2 \left (2 a^2-3 b^2\right ) \sin ^2(c+d x)-35 a b^3 \sin ^3(c+d x)\right )\right )}{3465 b^5 d} \end {gather*}

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 = 1.83, size = 126, normalized size = 0.82

method	result	size
default	\(\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} \left (315 b^{4} \left (\cos ^{4}\left (d x +c \right )\right )+280 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-240 a^{2} b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+360 b^{4} \left (\cos ^{2}\left (d x +c \right )\right )-192 a^{3} b \sin \left (d x +c \right )+512 a \,b^{3} \sin \left (d x +c \right )+128 a^{4}-288 a^{2} b^{2}+480 b^{4}\right )}{3465 b^{5} d}\)	\(126\)

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,method=_RETURNVERBOSE)

[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 = 0.28, size = 116, normalized size = 0.75 \begin {gather*} \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 {gather*}

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 = 0.36, size = 142, normalized size = 0.92 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 5.08, size = 161, normalized size = 1.05 \begin {gather*} \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 + 2970 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 2772 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 990 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} b^{2} + 2772 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a b^{2} - 2310 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 1155 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} b^{4}\right )}}{3465 \, b^{5} d} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^5\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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