3.877 \(\int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0415016, 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 = {18, 2748, 3767, 8, 3770} \[ \frac{A \sin (c+d x)}{b d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}+\frac{B \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0534019, size = 50, normalized size = 0.68 \[ \frac{\sqrt{\cos (c+d x)} \left (A \sin (c+d x)+B \cos (c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[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])^(3/2))

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Maple [A]  time = 0.315, 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 ) \sqrt{\cos \left ( dx+c \right ) } \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),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))*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2)

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Maxima [B]  time = 2.04411, size = 180, normalized size = 2.43 \begin{align*} \frac{\frac{4 \, A \sqrt{b} \sin \left (2 \, d x + 2 \, c\right )}{b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}} + \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{3}{2}}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/2*(4*A*sqrt(b)*sin(2*d*x + 2*c)/(b^2*cos(2*d*x + 2*c)^2 + b^2*sin(2*d*x + 2*c)^2 + 2*b^2*cos(2*d*x + 2*c) +
b^2) + 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^(3/2))/d

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Fricas [A]  time = 1.7188, size = 575, normalized size = 7.77 \begin{align*} \left [\frac{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^{2} d \cos \left (d x + c\right )^{2}}, -\frac{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} - \sqrt{b \cos \left (d x + c\right )} A \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} 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))/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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