3.860 \(\int \frac{(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=107 \[ \frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b^2 B x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b^2 B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0239772, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {17, 2734} \[ \frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{b^2 B x \sqrt{b \cos (c+d x)}}{2 \sqrt{\cos (c+d x)}}+\frac{b^2 B \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.123914, size = 57, normalized size = 0.53 \[ \frac{(b \cos (c+d x))^{5/2} (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*(4*A*Sin[c + d*x] + B*(2*(c + d*x) + Sin[2*(c + d*x)])))/(4*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 0.246, size = 55, normalized size = 0.5 \begin{align*}{\frac{B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +2\,A\sin \left ( dx+c \right ) +B \left ( dx+c \right ) }{2\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{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))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.03302, size = 63, normalized size = 0.59 \begin{align*} \frac{4 \, A b^{\frac{5}{2}} \sin \left (d x + c\right ) +{\left (2 \,{\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt{b}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*(4*A*b^(5/2)*sin(d*x + c) + (2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*B*sqrt(b))/d

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Fricas [A]  time = 1.7651, size = 597, normalized size = 5.58 \begin{align*} \left [\frac{B \sqrt{-b} b^{2} \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 (B b^{2} \cos \left (d x + c\right ) + 2 \, A b^{2}\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{B b^{\frac{5}{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 (B b^{2} \cos \left (d x + c\right ) + 2 \, A b^{2}\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))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(B*sqrt(-b)*b^2*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*(B*b^2*cos(d*x + c) + 2*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)), 1/2*(B*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x +
c) + (B*b^2*cos(d*x + c) + 2*A*b^2)*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))**(5/2)*(A+B*cos(d*x+c))/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 (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{5}{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))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(3/2), x)