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

Optimal. Leaf size=147 \[ \frac{b (3 A+2 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b B x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d}+\frac{b C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0590462, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.07, Rules used = {17, 3023, 2734} \[ \frac{b (3 A+2 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \sqrt{\cos (c+d x)}}+\frac{b B x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d}+\frac{b C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0720573, size = 76, normalized size = 0.52 \[ \frac{b \sqrt{b \cos (c+d x)} (3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x)))}{12 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[b*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[Cos[c + d*x]])

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Maple [A]  time = 0.418, size = 83, normalized size = 0.6 \begin{align*}{\frac{2\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +6\,A\sin \left ( dx+c \right ) +3\,B \left ( dx+c \right ) +4\,\sin \left ( dx+c \right ) C}{6\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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]

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

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Maxima [A]  time = 2.32264, size = 116, normalized size = 0.79 \begin{align*} \frac{12 \, A b^{\frac{3}{2}} \sin \left (d x + c\right ) + 3 \,{\left (2 \,{\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt{b} +{\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 )} C \sqrt{b}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

1/12*(12*A*b^(3/2)*sin(d*x + c) + 3*(2*(d*x + c)*b + b*sin(2*d*x + 2*c))*B*sqrt(b) + (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))))*C*sqrt(b))/d

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Fricas [A]  time = 1.96836, size = 684, normalized size = 4.65 \begin{align*} \left [\frac{3 \, B \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 (2 \, C b \cos \left (d x + c\right )^{2} + 3 \, B b \cos \left (d x + c\right ) + 2 \,{\left (3 \, A + 2 \, C\right )} b\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac{3 \, B 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 (2 \, C b \cos \left (d x + c\right )^{2} + 3 \, B b \cos \left (d x + c\right ) + 2 \,{\left (3 \, A + 2 \, C\right )} b\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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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((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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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