3.5.83 \(\int \cos ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) [483]

Optimal. Leaf size=154 \[ \frac {2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{11/2}}{11 b^5 d}+\frac {2 (a+b \sin (c+d x))^{13/2}}{13 b^5 d} \]

[Out]

2/5*(a^2-b^2)^2*(a+b*sin(d*x+c))^(5/2)/b^5/d-8/7*a*(a^2-b^2)*(a+b*sin(d*x+c))^(7/2)/b^5/d+4/9*(3*a^2-b^2)*(a+b
*sin(d*x+c))^(9/2)/b^5/d-8/11*a*(a+b*sin(d*x+c))^(11/2)/b^5/d+2/13*(a+b*sin(d*x+c))^(13/2)/b^5/d

________________________________________________________________________________________

Rubi [A]
time = 0.07, 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))^{9/2}}{9 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac {2 (a+b \sin (c+d x))^{13/2}}{13 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{11/2}}{11 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(2*(a^2 - b^2)^2*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d
) + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(9/2))/(9*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(11/2))/(11*b^5*d) + (2
*(a + b*Sin[c + d*x])^(13/2))/(13*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) (a+b \sin (c+d x))^{3/2} \, dx &=\frac {\text {Subst}\left (\int (a+x)^{3/2} \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 (a+x)^{3/2}-4 \left (a^3-a b^2\right ) (a+x)^{5/2}+2 \left (3 a^2-b^2\right ) (a+x)^{7/2}-4 a (a+x)^{9/2}+(a+x)^{11/2}\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))^{5/2}}{5 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{9/2}}{9 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{11/2}}{11 b^5 d}+\frac {2 (a+b \sin (c+d x))^{13/2}}{13 b^5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.99, size = 238, normalized size = 1.55 \begin {gather*} \frac {2 a \left (6144 a^6-35456 a^4 b^2+137910 a^2 b^4+29337 b^6\right ) \sqrt {1+\frac {b \sin (c+d x)}{a}} \left (-1+\sqrt {1+\frac {b \sin (c+d x)}{a}}\right )-b (a+b \sin (c+d x)) \left (b \left (2304 a^4-12048 a^2 b^2+35959 b^4\right ) \cos (2 (c+d x))+70 b^3 \left (-6 a^2+275 b^2\right ) \cos (4 (c+d x))+3465 b^5 \cos (6 (c+d x))+8 a \left (768 a^4-4216 a^2 b^2-40197 b^4\right ) \sin (c+d x)-20 a b^2 \left (48 a^2+3515 b^2\right ) \sin (3 (c+d x))-8820 a b^4 \sin (5 (c+d x))\right )}{720720 b^5 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(2*a*(6144*a^6 - 35456*a^4*b^2 + 137910*a^2*b^4 + 29337*b^6)*Sqrt[1 + (b*Sin[c + d*x])/a]*(-1 + Sqrt[1 + (b*Si
n[c + d*x])/a]) - b*(a + b*Sin[c + d*x])*(b*(2304*a^4 - 12048*a^2*b^2 + 35959*b^4)*Cos[2*(c + d*x)] + 70*b^3*(
-6*a^2 + 275*b^2)*Cos[4*(c + d*x)] + 3465*b^5*Cos[6*(c + d*x)] + 8*a*(768*a^4 - 4216*a^2*b^2 - 40197*b^4)*Sin[
c + d*x] - 20*a*b^2*(48*a^2 + 3515*b^2)*Sin[3*(c + d*x)] - 8820*a*b^4*Sin[5*(c + d*x)]))/(720720*b^5*d*Sqrt[a
+ b*Sin[c + d*x]])

________________________________________________________________________________________

Maple [A]
time = 1.58, size = 126, normalized size = 0.82

method	result	size
default	\(\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}} \left (3465 b^{4} \left (\cos ^{4}\left (d x +c \right )\right )+2520 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-1680 a^{2} b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+3080 b^{4} \left (\cos ^{2}\left (d x +c \right )\right )-960 a^{3} b \sin \left (d x +c \right )+3200 a \,b^{3} \sin \left (d x +c \right )+384 a^{4}-608 a^{2} b^{2}+2464 b^{4}\right )}{45045 b^{5} d}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/45045/b^5*(a+b*sin(d*x+c))^(5/2)*(3465*b^4*cos(d*x+c)^4+2520*a*b^3*cos(d*x+c)^2*sin(d*x+c)-1680*a^2*b^2*cos(
d*x+c)^2+3080*b^4*cos(d*x+c)^2-960*a^3*b*sin(d*x+c)+3200*a*b^3*sin(d*x+c)+384*a^4-608*a^2*b^2+2464*b^4)/d

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 116, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (3465 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 16380 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 10010 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 25740 \, {\left (a^{3} - a b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} + 9009 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\right )}}{45045 \, 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))^(3/2),x, algorithm="maxima")

[Out]

2/45045*(3465*(b*sin(d*x + c) + a)^(13/2) - 16380*(b*sin(d*x + c) + a)^(11/2)*a + 10010*(3*a^2 - b^2)*(b*sin(d
*x + c) + a)^(9/2) - 25740*(a^3 - a*b^2)*(b*sin(d*x + c) + a)^(7/2) + 9009*(a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x
+ c) + a)^(5/2))/(b^5*d)

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 184, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (3465 \, b^{6} \cos \left (d x + c\right )^{6} - 384 \, a^{6} + 2144 \, a^{4} b^{2} - 8256 \, a^{2} b^{4} - 2464 \, b^{6} - 35 \, {\left (3 \, a^{2} b^{4} + 11 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (18 \, a^{4} b^{2} - 81 \, a^{2} b^{4} - 77 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2205 \, a b^{5} \cos \left (d x + c\right )^{4} - 96 \, a^{5} b + 512 \, a^{3} b^{3} + 4064 \, a b^{5} + 20 \, {\left (3 \, a^{3} b^{3} + 137 \, a 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}}{45045 \, 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))^(3/2),x, algorithm="fricas")

[Out]

-2/45045*(3465*b^6*cos(d*x + c)^6 - 384*a^6 + 2144*a^4*b^2 - 8256*a^2*b^4 - 2464*b^6 - 35*(3*a^2*b^4 + 11*b^6)
*cos(d*x + c)^4 + 8*(18*a^4*b^2 - 81*a^2*b^4 - 77*b^6)*cos(d*x + c)^2 - 2*(2205*a*b^5*cos(d*x + c)^4 - 96*a^5*
b + 512*a^3*b^3 + 4064*a*b^5 + 20*(3*a^3*b^3 + 137*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) +
a)/(b^5*d)

________________________________________________________________________________________

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))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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))^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^5\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,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))^(3/2),x)

[Out]

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

________________________________________________________________________________________