\(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [298]

Optimal result

Integrand size = 43, antiderivative size = 189 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {b (4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {b B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]

[Out]

1/4*b*C*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/8*b*(4*A+3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(
1/2)+b*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*b*B*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/cos(d*
x+c)^(1/2)+1/8*b*(4*A+3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {17, 3102, 2827, 2715, 8, 2713} \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {b x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b (4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}-\frac {b B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

[In]

Int[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(b*(4*A + 3*C)*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (b*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqr
t[Cos[c + d*x]]) + (b*(4*A + 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (b*C*Cos[c + d
*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d) - (b*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Sqrt[Cos[c
+ d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3102

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {b C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) (4 A+3 C+4 B \cos (c+d x)) \, dx}{4 \sqrt {\cos (c+d x)}} \\ & = \frac {b C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left (b B \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{\sqrt {\cos (c+d x)}}+\frac {\left (b (4 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}} \\ & = \frac {b (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left (b (4 A+3 C) \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt {\cos (c+d x)}}-\frac {\left (b B \sqrt {b \cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}} \\ & = \frac {b (4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {b B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.49 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{3/2} (48 A c+36 c C+48 A d x+36 C d x+72 B \sin (c+d x)+24 (A+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Integrate[Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

((b*Cos[c + d*x])^(3/2)*(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 72*B*Sin[c + d*x] + 24*(A + C)*Sin[2*(c + d*x
)] + 8*B*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d*Cos[c + d*x]^(3/2))

Maple [A] (verified)

Time = 8.63 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (6 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+12 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+9 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+12 A \left (d x +c \right )+16 B \sin \left (d x +c \right )+9 C \left (d x +c \right )\right )}{24 d \sqrt {\cos \left (d x +c \right )}}\)	\(115\)
parts	\(\frac {A b \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {B b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {C b \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\)	\(149\)
risch	\(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x \left (8 A +6 C \right )}{8 \,{\mathrm e}^{2 i \left (d x +c \right )}+8}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} C}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{4 i \left (d x +c \right )} B}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {3 i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {3 i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} \left (A +C \right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} B}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (8 A +7 C \right ) \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (9 C +8 A \right ) \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(441\)

[In]

int((cos(d*x+c)*b)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*b/d*(cos(d*x+c)*b)^(1/2)*(6*C*cos(d*x+c)^3*sin(d*x+c)+8*B*sin(d*x+c)*cos(d*x+c)^2+12*A*sin(d*x+c)*cos(d*x
+c)+9*C*cos(d*x+c)*sin(d*x+c)+12*A*(d*x+c)+16*B*sin(d*x+c)+9*C*(d*x+c))/cos(d*x+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.51 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {-b} b \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, C b \cos \left (d x + c\right )^{3} + 8 \, B b \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right ) + 16 \, B b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {3 \, {\left (4 \, A + 3 \, C\right )} b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, C b \cos \left (d x + c\right )^{3} + 8 \, B b \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right ) + 16 \, B b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*(4*A + 3*C)*sqrt(-b)*b*cos(d*x + c)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos
(d*x + c))*sin(d*x + c) - b) + 2*(6*C*b*cos(d*x + c)^3 + 8*B*b*cos(d*x + c)^2 + 3*(4*A + 3*C)*b*cos(d*x + c) +
16*B*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/24*(3*(4*A + 3*C)*b^(3/2)*a
rctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (6*C*b*cos(d*x + c)^3 + 8
*B*b*cos(d*x + c)^2 + 3*(4*A + 3*C)*b*cos(d*x + c) + 16*B*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.67 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (2 \, {\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + 8 \, {\left (b \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b} + 3 \, {\left (12 \, {\left (d x + c\right )} b + b \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{96 \, d} \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/96*(24*(2*(d*x + c)*b + b*sin(2*d*x + 2*c))*A*sqrt(b) + 8*(b*sin(3*d*x + 3*c) + 9*b*sin(1/3*arctan2(sin(3*d*
x + 3*c), cos(3*d*x + 3*c))))*B*sqrt(b) + 3*(12*(d*x + c)*b + b*sin(4*d*x + 4*c) + 8*b*sin(1/2*arctan2(sin(4*d
*x + 4*c), cos(4*d*x + 4*c))))*C*sqrt(b))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (161) = 322\).

Time = 4.13 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.27 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (12 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 9 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 48 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 36 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 54 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A \sqrt {b} d x + 9 \, C \sqrt {b} d x + 24 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} b}{24 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/24*(12*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^8 + 9*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^8 + 48*A*sqrt(b)*d*x*tan(
1/2*d*x + 1/2*c)^6 + 36*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^6 - 24*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^7 + 48*B*sqrt
(b)*tan(1/2*d*x + 1/2*c)^7 - 30*C*sqrt(b)*tan(1/2*d*x + 1/2*c)^7 + 72*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 + 5
4*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 - 24*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 80*B*sqrt(b)*tan(1/2*d*x + 1/2*
c)^5 + 18*C*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 48*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^2 + 36*C*sqrt(b)*d*x*tan(1/
2*d*x + 1/2*c)^2 + 24*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^3 + 80*B*sqrt(b)*tan(1/2*d*x + 1/2*c)^3 - 18*C*sqrt(b)*ta
n(1/2*d*x + 1/2*c)^3 + 12*A*sqrt(b)*d*x + 9*C*sqrt(b)*d*x + 24*A*sqrt(b)*tan(1/2*d*x + 1/2*c) + 48*B*sqrt(b)*t
an(1/2*d*x + 1/2*c) + 30*C*sqrt(b)*tan(1/2*d*x + 1/2*c))*b/(d*tan(1/2*d*x + 1/2*c)^8 + 4*d*tan(1/2*d*x + 1/2*c
)^6 + 6*d*tan(1/2*d*x + 1/2*c)^4 + 4*d*tan(1/2*d*x + 1/2*c)^2 + d)

Mupad [B] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.73 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {b\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (24\,A\,\sin \left (c+d\,x\right )+24\,C\,\sin \left (c+d\,x\right )+24\,A\,\sin \left (3\,c+3\,d\,x\right )+80\,B\,\sin \left (2\,c+2\,d\,x\right )+8\,B\,\sin \left (4\,c+4\,d\,x\right )+27\,C\,\sin \left (3\,c+3\,d\,x\right )+3\,C\,\sin \left (5\,c+5\,d\,x\right )+96\,A\,d\,x\,\cos \left (c+d\,x\right )+72\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{96\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

(b*cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(24*A*sin(c + d*x) + 24*C*sin(c + d*x) + 24*A*sin(3*c + 3*d*x) +
80*B*sin(2*c + 2*d*x) + 8*B*sin(4*c + 4*d*x) + 27*C*sin(3*c + 3*d*x) + 3*C*sin(5*c + 5*d*x) + 96*A*d*x*cos(c +
d*x) + 72*C*d*x*cos(c + d*x)))/(96*d*(cos(2*c + 2*d*x) + 1))