3.114 \(\int \frac{(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=85 \[ \frac{b^2 (2 A+3 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.0520294, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {17, 3012, 3767, 8} \[ \frac{b^2 (2 A+3 C) \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{A b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{3 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

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

Rubi steps

\begin{align*} \int \frac{(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{13}{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 ^4(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{\left (b^2 (2 A+3 C) \sqrt{b \cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{3 \sqrt{\cos (c+d x)}}\\ &=\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{\left (b^2 (2 A+3 C) \sqrt{b \cos (c+d x)}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d \sqrt{\cos (c+d x)}}\\ &=\frac{A b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{b^2 (2 A+3 C) \sqrt{b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.217753, size = 51, normalized size = 0.6 \[ \frac{\sin (c+d x) (b \cos (c+d x))^{5/2} \left (A \tan ^2(c+d x)+3 (A+C)\right )}{3 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*Sin[c + d*x]*(3*(A + C) + A*Tan[c + d*x]^2))/(3*d*Cos[c + d*x]^(7/2))

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Maple [A]  time = 0.42, size = 54, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+A \right ) \sin \left ( dx+c \right ) }{3\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{11}{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)^(13/2),x)

[Out]

1/3/d*(2*A*cos(d*x+c)^2+3*C*cos(d*x+c)^2+A)*sin(d*x+c)*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2)

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Maxima [B]  time = 2.12351, size = 495, normalized size = 5.82 \begin{align*} \frac{2 \,{\left (\frac{3 \, C 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} - \frac{2 \,{\left (3 \, b^{2} \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) -{\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (6 \, d x + 6 \, c\right ) - 3 \,{\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )\right )} A \sqrt{b}}{2 \,{\left (3 \, \cos \left (4 \, d x + 4 \, c\right ) + 3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (6 \, d x + 6 \, c\right ) + \cos \left (6 \, d x + 6 \, c\right )^{2} + 6 \,{\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 9 \, \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 6 \,{\left (\sin \left (4 \, d x + 4 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right ) + \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{3 \, 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)^(13/2),x, algorithm="maxima")

[Out]

2/3*(3*C*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) - 2*(3*b^
2*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) + 9*b^2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c) - (3*b^2*cos(2*d*x + 2*c) + b^2)
*sin(6*d*x + 6*c) - 3*(3*b^2*cos(2*d*x + 2*c) + b^2)*sin(4*d*x + 4*c))*A*sqrt(b)/(2*(3*cos(4*d*x + 4*c) + 3*co
s(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*co
s(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x
+ 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x +
2*c) + 1))/d

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Fricas [A]  time = 1.44071, size = 139, normalized size = 1.64 \begin{align*} \frac{{\left ({\left (2 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + A b^{2}\right )} \sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{\frac{7}{2}}} \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)^(13/2),x, algorithm="fricas")

[Out]

1/3*((2*A + 3*C)*b^2*cos(d*x + c)^2 + A*b^2)*sqrt(b*cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^(7/2))

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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+C*cos(d*x+c)**2)/cos(d*x+c)**(13/2),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{13}{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)^(13/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(13/2), x)