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

Optimal. Leaf size=314 \[ \frac{2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b^2 d}+\frac{2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^2 d}-\frac{2 a \left (31 a^2 b^2+8 a^4-39 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (33 a^2 b^2+8 a^4+147 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d} \]

[Out]

(2*(8*a^4 + 33*a^2*b^2 + 147*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(315*b^3*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(8*a^4 + 31*a^2*b^2 - 39*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ell
ipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*(8*a^2 + 39*b^2)*Sqrt[a + b*Co
s[c + d*x]]*Sin[c + d*x])/(315*b^2*d) + (2*(8*a^2 + 49*b^2)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b^2*
d) - (8*a*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*b^2*d) + (2*Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2)*Sin
[c + d*x])/(9*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.517743, antiderivative size = 314, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2793, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (8 a^2+49 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b^2 d}+\frac{2 a \left (8 a^2+39 b^2\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b^2 d}-\frac{2 a \left (31 a^2 b^2+8 a^4-39 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (33 a^2 b^2+8 a^4+147 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) (a+b \cos (c+d x))^{5/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(8*a^4 + 33*a^2*b^2 + 147*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(315*b^3*d*S
qrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(8*a^4 + 31*a^2*b^2 - 39*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ell
ipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^3*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*(8*a^2 + 39*b^2)*Sqrt[a + b*Co
s[c + d*x]]*Sin[c + d*x])/(315*b^2*d) + (2*(8*a^2 + 49*b^2)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b^2*
d) - (8*a*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*b^2*d) + (2*Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2)*Sin
[c + d*x])/(9*b*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \cos (c+d x))^{3/2} \, dx &=\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}+\frac{2 \int (a+b \cos (c+d x))^{3/2} \left (a+\frac{7}{2} b \cos (c+d x)-2 a \cos ^2(c+d x)\right ) \, dx}{9 b}\\ &=-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}+\frac{4 \int (a+b \cos (c+d x))^{3/2} \left (-\frac{3 a b}{2}+\frac{1}{4} \left (8 a^2+49 b^2\right ) \cos (c+d x)\right ) \, dx}{63 b^2}\\ &=\frac{2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}+\frac{8 \int \sqrt{a+b \cos (c+d x)} \left (-\frac{3}{8} b \left (2 a^2-49 b^2\right )+\frac{3}{8} a \left (8 a^2+39 b^2\right ) \cos (c+d x)\right ) \, dx}{315 b^2}\\ &=\frac{2 a \left (8 a^2+39 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^2 d}+\frac{2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}+\frac{16 \int \frac{\frac{3}{8} a b \left (a^2+93 b^2\right )+\frac{3}{16} \left (8 a^4+33 a^2 b^2+147 b^4\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{945 b^2}\\ &=\frac{2 a \left (8 a^2+39 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^2 d}+\frac{2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}-\frac{\left (a \left (8 a^4+31 a^2 b^2-39 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^3}+\frac{\left (8 a^4+33 a^2 b^2+147 b^4\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{315 b^3}\\ &=\frac{2 a \left (8 a^2+39 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^2 d}+\frac{2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}+\frac{\left (\left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{315 b^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{315 b^3 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 a \left (8 a^4+31 a^2 b^2-39 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (8 a^2+39 b^2\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b^2 d}+\frac{2 \left (8 a^2+49 b^2\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b^2 d}-\frac{8 a (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b^2 d}+\frac{2 \cos (c+d x) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{9 b d}\\ \end{align*}

Mathematica [A]  time = 1.38139, size = 262, normalized size = 0.83 \[ \frac{b \sin (c+d x) \left (\left (1606 a b^3-8 a^3 b\right ) \cos (c+d x)+4 \left (53 a^2 b^2+84 b^4\right ) \cos (2 (c+d x))+916 a^2 b^2-32 a^4+170 a b^3 \cos (3 (c+d x))+35 b^4 \cos (4 (c+d x))+301 b^4\right )-8 a \left (31 a^2 b^2+8 a^4-39 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+8 \left (33 a^3 b^2+33 a^2 b^3+8 a^4 b+8 a^5+147 a b^4+147 b^5\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{1260 b^3 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(8*a^5 + 8*a^4*b + 33*a^3*b^2 + 33*a^2*b^3 + 147*a*b^4 + 147*b^5)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipt
icE[(c + d*x)/2, (2*b)/(a + b)] - 8*a*(8*a^4 + 31*a^2*b^2 - 39*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*b)/(a + b)] + b*(-32*a^4 + 916*a^2*b^2 + 301*b^4 + (-8*a^3*b + 1606*a*b^3)*Cos[c + d*x] + 4
*(53*a^2*b^2 + 84*b^4)*Cos[2*(c + d*x)] + 170*a*b^3*Cos[3*(c + d*x)] + 35*b^4*Cos[4*(c + d*x)])*Sin[c + d*x])/
(1260*b^3*d*Sqrt[a + b*Cos[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 3.608, size = 995, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*b^5*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^10+(1360*a*b^4+2240*b^5)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-424*a^2*b^3-2040*a*b^4-2072*b^5)*sin(
1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(-4*a^3*b^2+424*a^2*b^3+1568*a*b^4+952*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d
*x+1/2*c)+(8*a^4*b+2*a^3*b^2-282*a^2*b^3-444*a*b^4-168*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-8*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a
-b))^(1/2))*a^5-31*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2+39*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^
2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^4+8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2
*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5-8*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/
(a-b))^(1/2))*a^4*b+33*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b^2-33*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2
*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^3+147*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^
4-147*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),(-2*b/(a-b))^(1/2))*b^5)/b^3/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/
2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{3}\right )} \sqrt{b \cos \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^4 + a*cos(d*x + c)^3)*sqrt(b*cos(d*x + c) + a), x)

________________________________________________________________________________________

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*cos(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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