3.863 \(\int \frac{(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=74 \[ \frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b^2 B \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0426741, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {17, 2748, 3767, 8, 3770} \[ \frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b^2 B \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

(b^2*B*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(d*Sqrt[Cos[c + d*x]]) + (A*b^2*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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{\left (A b^2 \sqrt{b \cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{\sqrt{\cos (c+d x)}}+\frac{\left (b^2 B \sqrt{b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{b^2 B \tanh ^{-1}(\sin (c+d x)) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}-\frac{\left (A b^2 \sqrt{b \cos (c+d x)}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d \sqrt{\cos (c+d x)}}\\ &=\frac{b^2 B \tanh ^{-1}(\sin (c+d x)) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}+\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.0784006, size = 50, normalized size = 0.68 \[ \frac{(b \cos (c+d x))^{5/2} \left (A \sin (c+d x)+B \cos (c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*(B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x] + A*Sin[c + d*x]))/(d*Cos[c + d*x]^(7/2))

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Maple [A]  time = 0.224, size = 59, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -2\,B\cos \left ( dx+c \right ){\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +A\sin \left ( dx+c \right ) \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{7}{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)^(9/2),x)

[Out]

1/d*(-2*B*cos(d*x+c)*arctanh((-1+cos(d*x+c))/sin(d*x+c))+A*sin(d*x+c))*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2)

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Maxima [A]  time = 1.9862, size = 171, normalized size = 2.31 \begin{align*} \frac{\frac{4 \, A b^{\frac{5}{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} +{\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} B \sqrt{b}}{2 \, 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)^(9/2),x, algorithm="maxima")

[Out]

1/2*(4*A*b^(5/2)*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) + (b^2*lo
g(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x
+ c) + 1))*B*sqrt(b))/d

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Fricas [A]  time = 1.62102, size = 581, normalized size = 7.85 \begin{align*} \left [\frac{B b^{\frac{5}{2}} \cos \left (d x + c\right )^{2} \log \left (-\frac{b \cos \left (d x + c\right )^{3} - 2 \, \sqrt{b \cos \left (d x + c\right )} \sqrt{b} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt{b \cos \left (d x + c\right )} A b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}, -\frac{B \sqrt{-b} b^{2} \arctan \left (\frac{\sqrt{b \cos \left (d x + c\right )} \sqrt{-b} \sin \left (d x + c\right )}{b \sqrt{\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - \sqrt{b \cos \left (d x + c\right )} A b^{2} \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))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/2*(B*b^(5/2)*cos(d*x + c)^2*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(
d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(
d*cos(d*x + c)^2), -(B*sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*
cos(d*x + c)^2 - sqrt(b*cos(d*x + c))*A*b^2*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))**(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(9/2),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{9}{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)^(9/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(9/2), x)