3.103 \(\int \frac{(b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{A b \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b C x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0322901, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {17, 3012, 8} \[ \frac{A b \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b C x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

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

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 3012

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

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+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{\left (b \sqrt{b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A b \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (b C \sqrt{b \cos (c+d x)}\right ) \int 1 \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{b C x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{A b \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.0595929, size = 45, normalized size = 0.74 \[ \frac{(b \cos (c+d x))^{3/2} (A \sin (c+d x)+C d x \cos (c+d x))}{d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

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

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Maple [A]  time = 0.228, size = 45, normalized size = 0.7 \begin{align*}{\frac{C\cos \left ( dx+c \right ) \left ( dx+c \right ) +A\sin \left ( dx+c \right ) }{d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)

[Out]

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

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Maxima [A]  time = 1.82936, size = 108, normalized size = 1.77 \begin{align*} \frac{2 \,{\left (C b^{\frac{3}{2}} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + \frac{A b^{\frac{3}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.61463, size = 527, normalized size = 8.64 \begin{align*} \left [\frac{C \sqrt{-b} b \cos \left (d x + c\right )^{2} \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 \, \sqrt{b \cos \left (d x + c\right )} A b \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}, \frac{C 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 )^{2} + \sqrt{b \cos \left (d x + c\right )} A b \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

[1/2*(C*sqrt(-b)*b*cos(d*x + c)^2*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*sqrt(b*cos(d*x + c))*A*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2), (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)^2 + sqrt(b*cos(d*x + c))*
A*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)]

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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+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),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} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\cos \left (d x + c\right )^{\frac{7}{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+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)/cos(d*x + c)^(7/2), x)