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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 = {17, 2748, 3767, 8, 3770} \[ \frac {A \sin (c+d x)}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b^2 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\left (A \sqrt {\cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{b^2 \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 {B \tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b^2 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^2 d \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*}

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Mathematica [A]  time = 0.05, size = 50, normalized size = 0.68 \[ \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A \sin (c+d x)+B \cos (c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(3/2)*(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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fricas [A]  time = 0.92, size = 211, normalized size = 2.85 \[ \left [\frac {B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}, -\frac {B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{3} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),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^3*
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^3*d*cos(d*x + c)^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 59, normalized size = 0.80 \[ \frac {\left (-2 B \cos \left (d x +c \right ) \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+A \sin \left (d x +c \right )\right ) \left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.66, size = 133, normalized size = 1.80 \[ \frac {\frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}} + \frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {5}{2}}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(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))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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