3.111 \(\int \frac{(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0369129, antiderivative size = 74, 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, 3014, 3770} \[ \frac{A b^2 \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\cos (c+d x)}}+\frac{b^2 C \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

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

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 3014

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

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

Mathematica [A]  time = 0.0802229, size = 44, normalized size = 0.59 \[ \frac{(b \cos (c+d x))^{5/2} \left (A \tanh ^{-1}(\sin (c+d x))+C \sin (c+d x)\right )}{d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

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

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Maple [A]  time = 0.221, size = 55, normalized size = 0.7 \begin{align*} -{\frac{1}{d} \left ( 2\,A{\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -\sin \left ( dx+c \right ) C \right ) \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+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)

[Out]

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

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Maxima [A]  time = 2.08183, size = 117, normalized size = 1.58 \begin{align*} \frac{2 \, C b^{\frac{5}{2}} \sin \left (d x + c\right ) +{\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 )} A \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+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75235, size = 570, normalized size = 7.7 \begin{align*} \left [\frac{A b^{\frac{5}{2}} \cos \left (d x + c\right ) \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 )} C b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, -\frac{A \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 ) - \sqrt{b \cos \left (d x + c\right )} C b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{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+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

[1/2*(A*b^(5/2)*cos(d*x + c)*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))*C*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)), -(A*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) - sqrt(b*cos(d*x + c))*C*b^2*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+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{5}{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))^(5/2)*(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))^(5/2)/cos(d*x + c)^(7/2), x)