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

Optimal. Leaf size=102 \[ \frac{A \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{B \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b^2 d \sqrt{b \cos (c+d x)}}+\frac{C x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0621749, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.093, Rules used = {17, 3021, 2735, 3770} \[ \frac{A \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{B \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b^2 d \sqrt{b \cos (c+d x)}}+\frac{C x \sqrt{\cos (c+d x)}}{b^2 \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x*Sqrt[Cos[c + d*x]])/(b^2*Sqrt[b*Cos[c + d*x]]) + (B*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(b^2*d*Sqrt
[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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

Mathematica [A]  time = 0.0691534, size = 60, normalized size = 0.59 \[ \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A \sin (c+d x)+B \cos (c+d x) \tanh ^{-1}(\sin (c+d x))+C d x \cos (c+d x)\right )}{d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.384, size = 72, normalized size = 0.7 \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 ) +C\cos \left ( dx+c \right ) \left ( dx+c \right ) +A\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.08213, size = 212, normalized size = 2.08 \begin{align*} \frac{\frac{4 \, A \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{B{\left (\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 ) - \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^{\frac{5}{2}}} + \frac{4 \, C \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac{5}{2}}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.19824, size = 886, normalized size = 8.69 \begin{align*} \left [-\frac{2 \, B \sqrt{-b} \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} + C \sqrt{-b} \cos \left (d x + c\right )^{2} \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 \, \sqrt{b \cos \left (d x + c\right )} A \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}, \frac{2 \, C \sqrt{b} \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 )^{2} + B \sqrt{b} \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 \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/2*(2*B*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^2 +
C*sqrt(-b)*cos(d*x + c)^2*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*sqrt(b*cos(d*x + c))*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^2), 1/2*(2*C*sqrt(b
)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c)^2 + B*sqrt(b)*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*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2)/(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{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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