\(\int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [90]

Optimal result

Integrand size = 35, antiderivative size = 113 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d} \]

[Out]

1/4*C*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/8*(4*A+3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)
+1/8*(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 = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {17, 3093, 2715, 8} \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {(4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

[In]

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

[Out]

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

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 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 3093

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

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.59 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b \cos (c+d x)} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[b*Cos[c + d*x]]*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + C*Sin[4*(c + d*x)]))/(32*d*Sqrt[
Cos[c + d*x]])

Maple [A] (verified)

Time = 7.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (2 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+4 A \left (d x +c \right )+3 C \left (d x +c \right )\right )}{8 d \sqrt {\cos \left (d x +c \right )}}\)	\(88\)
parts	\(\frac {A \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\)	\(106\)
risch	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x \left (8 A +6 C \right )}{8 \,{\mathrm e}^{2 i \left (d x +c \right )}+8}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} C}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} \left (A +C \right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (8 A +7 C \right ) \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (9 C +8 A \right ) \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(253\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.77 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\left [\frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (4 \, A + 3 \, C\right )} \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{16 \, d}, \frac {{\left (2 \, C \cos \left (d x + c\right )^{2} + 4 \, A + 3 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (4 \, A + 3 \, C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{8 \, d}\right ] \]

[In]

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

[Out]

[1/16*(2*(2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (4*A + 3*C)*s
qrt(-b)*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))/d, 1/8*
((2*C*cos(d*x + c)^2 + 4*A + 3*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (4*A + 3*C)*sqrt(b)*a
rctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))))/d]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{32 \, d} \]

[In]

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

[Out]

1/32*(8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*sqrt(b) + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*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. 362 vs. \(2 (95) = 190\).

Time = 2.96 (sec) , antiderivative size = 362, normalized size of antiderivative = 3.20 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {4 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 3 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 16 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 8 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 10 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, C \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A \sqrt {b} d x + 3 \, C \sqrt {b} d x + 8 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, C \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{8 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]

[In]

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

[Out]

1/8*(4*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^8 + 3*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^8 + 16*A*sqrt(b)*d*x*tan(1/
2*d*x + 1/2*c)^6 + 12*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^6 - 8*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^7 - 10*C*sqrt(b)
*tan(1/2*d*x + 1/2*c)^7 + 24*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 + 18*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 -
8*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 6*C*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 16*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c
)^2 + 12*C*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^2 + 8*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^3 - 6*C*sqrt(b)*tan(1/2*d*x +
1/2*c)^3 + 4*A*sqrt(b)*d*x + 3*C*sqrt(b)*d*x + 8*A*sqrt(b)*tan(1/2*d*x + 1/2*c) + 10*C*sqrt(b)*tan(1/2*d*x +
1/2*c))/(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 = 2.51 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99 \[ \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,A\,\sin \left (c+d\,x\right )+8\,C\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (3\,c+3\,d\,x\right )+9\,C\,\sin \left (3\,c+3\,d\,x\right )+C\,\sin \left (5\,c+5\,d\,x\right )+32\,A\,d\,x\,\cos \left (c+d\,x\right )+24\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(8*A*sin(c + d*x) + 8*C*sin(c + d*x) + 8*A*sin(3*c + 3*d*x) + 9*C*s
in(3*c + 3*d*x) + C*sin(5*c + 5*d*x) + 32*A*d*x*cos(c + d*x) + 24*C*d*x*cos(c + d*x)))/(32*d*(cos(2*c + 2*d*x)
+ 1))