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

Optimal. Leaf size=85 \[ \frac{(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{3 b^2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0469938, 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 = {18, 3012, 3767, 8} \[ \frac{(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{3 b^2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[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{A+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{A \sin (c+d x)}{3 b^2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\left ((2 A+3 C) \sqrt{\cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{3 b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{A \sin (c+d x)}{3 b^2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}-\frac{\left ((2 A+3 C) \sqrt{\cos (c+d x)}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 b^2 d \sqrt{b \cos (c+d x)}}\\ &=\frac{A \sin (c+d x)}{3 b^2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.111845, size = 51, normalized size = 0.6 \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A \tan ^2(c+d x)+3 (A+C)\right )}{3 d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.256, size = 54, normalized size = 0.6 \begin{align*}{\frac{\sin \left ( dx+c \right ) \left ( 2\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+A \right ) }{3\,d} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.93057, size = 556, normalized size = 6.54 \begin{align*} \frac{2 \,{\left (\frac{3 \, C \sqrt{b} \sin \left (2 \, d x + 2 \, c\right )}{b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}} + \frac{2 \,{\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \,{\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A}{{\left (b^{2} \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \,{\left (3 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \,{\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right ) + 6 \,{\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right )\right )} \sqrt{b}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.34058, size = 134, normalized size = 1.58 \begin{align*} \frac{{\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + A\right )} \sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b^{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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{\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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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