\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\) [316]

Optimal result

Integrand size = 43, antiderivative size = 143 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {(3 A+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {17, 3102, 2813} \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(2*Sqrt[b*Cos[c + d*x]]) + ((3*A + 2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[b*
Cos[c + d*x]]) + (B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[b*Cos[c + d*x]]) + (C*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(3*d*Sqrt[b*Cos[c + d*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 2813

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

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 {\sqrt {\cos (c+d x)} \int \cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {b \cos (c+d x)}} \\ & = \frac {C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x)) \, dx}{3 \sqrt {b \cos (c+d x)}} \\ & = \frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {(3 A+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {b \cos (c+d x)}}+\frac {B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} (6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x)))}{12 d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)])
)/(12*d*Sqrt[b*Cos[c + d*x]])

Maple [A] (verified)

Time = 9.46 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 A \sin \left (d x +c \right )+3 B \left (d x +c \right )+4 \sin \left (d x +c \right ) C \right )}{6 d \sqrt {\cos \left (d x +c \right ) b}}\)	\(83\)
parts	\(\frac {A \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}}+\frac {B \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{3 d \sqrt {\cos \left (d x +c \right ) b}}\)	\(113\)
risch	\(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(126\)

[In]

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

[Out]

1/6/d*cos(d*x+c)^(1/2)*(2*C*cos(d*x+c)^2*sin(d*x+c)+3*B*sin(d*x+c)*cos(d*x+c)+6*A*sin(d*x+c)+3*B*(d*x+c)+4*sin
(d*x+c)*C)/(cos(d*x+c)*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\left [-\frac {3 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b d \cos \left (d x + c\right )}, \frac {3 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[-1/12*(3*B*sqrt(-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*(2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c))/(b*d*cos(d*x + c)), 1/6*(3*B*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d
*x + c)^(3/2)))*cos(d*x + c) + (2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(c
os(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{\sqrt {b}} + \frac {C {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{\sqrt {b}} + \frac {12 \, A \sin \left (d x + c\right )}{\sqrt {b}}}{12 \, d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (3\,B\,\sin \left (c+d\,x\right )+12\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,B\,\sin \left (3\,c+3\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )+12\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{12\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x) + 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)*(3*B*sin(c + d*x) + 12*A*sin(2*c + 2*d*x) + 3*B*sin(3*c + 3*d*x) +
10*C*sin(2*c + 2*d*x) + C*sin(4*c + 4*d*x) + 12*B*d*x*cos(c + d*x)))/(12*b*d*(cos(2*c + 2*d*x) + 1))