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

Optimal. Leaf size=127 \[ \frac{A b x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b C x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b C \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0381749, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.093, Rules used = {17, 2637, 2635, 8} \[ \frac{A b x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b C x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b C \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d} \]

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))/Cos[c + d*x]^(3/2),x]

[Out]

(A*b*x*Sqrt[b*Cos[c + d*x]])/Sqrt[Cos[c + d*x]] + (b*C*x*Sqrt[b*Cos[c + d*x]])/(2*Sqrt[Cos[c + d*x]]) + (b*B*S
qrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (b*C*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c
+ d*x])/(2*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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

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 )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{\left (b \sqrt{b \cos (c+d x)}\right ) \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A b x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{\left (b B \sqrt{b \cos (c+d x)}\right ) \int \cos (c+d x) \, dx}{\sqrt{\cos (c+d x)}}+\frac{\left (b C \sqrt{b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A b x \sqrt{b \cos (c+d x)}}{\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 C \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac{\left (b C \sqrt{b \cos (c+d x)}\right ) \int 1 \, dx}{2 \sqrt{\cos (c+d x)}}\\ &=\frac{A b x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{b C x \sqrt{b \cos (c+d x)}}{2 \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 C \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)} \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.123911, size = 61, normalized size = 0.48 \[ \frac{(b \cos (c+d x))^{3/2} (2 (2 A+C) (c+d x)+4 B \sin (c+d x)+C \sin (2 (c+d x)))}{4 d \cos ^{\frac{3}{2}}(c+d x)} \]

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))/Cos[c + d*x]^(3/2),x]

[Out]

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

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Maple [A]  time = 0.283, size = 63, normalized size = 0.5 \begin{align*}{\frac{C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +2\,A \left ( dx+c \right ) +2\,B\sin \left ( dx+c \right ) +C \left ( dx+c \right ) }{2\,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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.09056, size = 90, normalized size = 0.71 \begin{align*} \frac{8 \, A b^{\frac{3}{2}} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 4 \, B b^{\frac{3}{2}} \sin \left (d x + c\right ) +{\left (2 \,{\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} C \sqrt{b}}{4 \, 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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.99187, size = 605, normalized size = 4.76 \begin{align*} \left [\frac{{\left (2 \, A + 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 (C b \cos \left (d x + c\right ) + 2 \, B b\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )}, \frac{{\left (2 \, A + 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 (C b \cos \left (d x + c\right ) + 2 \, B b\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, 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)^(3/2),x, algorithm="fricas")

[Out]

[1/4*((2*A + 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*(C*b*cos(d*x + c) + 2*B*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(
d*cos(d*x + c)), 1/2*((2*A + C)*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) + (C*b*cos(d*x + c) + 2*B*b)*sqrt(b*cos(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)**(3/2),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}}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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