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

Optimal. Leaf size=143 \[ \frac {(3 A+2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {b \cos (c+d x)}} \]

[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)

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Rubi [A]  time = 0.06, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {17, 3023, 2734} \[ \frac {(3 A+2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[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 2734

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 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]

Rubi steps

\begin {align*} \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)} \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*}

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Mathematica [A]  time = 0.20, size = 75, normalized size = 0.52 \[ \frac {\sqrt {\cos (c+d x)} (3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x)))}{12 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 1.72, size = 242, normalized size = 1.69 \[ \left [-\frac {3 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b d \cos \left (d x + c\right )}, \frac {3 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {b \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.41, size = 83, normalized size = 0.58 \[ \frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 A \sin \left (d x +c \right )+3 B \left (d x +c \right )+4 C \sin \left (d x +c \right )\right )}{6 d \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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]

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

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maxima [A]  time = 0.70, size = 80, normalized size = 0.56 \[ \frac {\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{\sqrt {b}} + \frac {C {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{\sqrt {b}} + \frac {12 \, A \sin \left (d x + c\right )}{\sqrt {b}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.41, size = 107, normalized size = 0.75 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (3\,B\,\sin \left (c+d\,x\right )+12\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,B\,\sin \left (3\,c+3\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )+12\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{12\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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