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

Optimal. Leaf size=107 \[ \frac{A \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{A \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0545298, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {18, 2748, 3768, 3770, 3767, 8} \[ \frac{A \sin (c+d x)}{2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{A \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(2*d*Sqrt[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(2*d*Cos[c + d*x]^(
3/2)*Sqrt[b*Cos[c + d*x]]) + (B*Sin[c + d*x])/(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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

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

Mathematica [A]  time = 0.0643787, size = 65, normalized size = 0.61 \[ \frac{\sin (c+d x) (A+2 B \cos (c+d x))+A \cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))}{2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.338, size = 120, normalized size = 1.1 \begin{align*}{\frac{1}{2\,d} \left ( -A\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+A\ln \left ({\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +A\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{b\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(-A*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+A*ln((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))*c
os(d*x+c)^2+2*B*sin(d*x+c)*cos(d*x+c)+A*sin(d*x+c))/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)

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Maxima [B]  time = 2.02027, size = 975, normalized size = 9.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*B*sqrt(b)*sin(2*d*x + 2*c)/(b*cos(2*d*x + 2*c)^2 + b*sin(2*d*x + 2*c)^2 + 2*b*cos(2*d*x + 2*c) + b) - (
4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x
+ 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*c) + 1
)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2
*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2
+ sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*lo
g(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
)^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) +
1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(1
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A/((2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x +
4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)
^2 + 4*cos(2*d*x + 2*c) + 1)*sqrt(b)))/d

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Fricas [A]  time = 1.6147, size = 632, normalized size = 5.91 \begin{align*} \left [\frac{A \sqrt{b} \cos \left (d x + c\right )^{3} \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 \,{\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )^{3}}, -\frac{A \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 )^{3} -{\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{3}}\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)^(5/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/4*(A*sqrt(b)*cos(d*x + c)^3*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*(2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c)
)*sin(d*x + c))/(b*d*cos(d*x + c)^3), -1/2*(A*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sq
rt(cos(d*x + c))))*cos(d*x + c)^3 - (2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c
))/(b*d*cos(d*x + c)^3)]

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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)**(5/2)/(b*cos(d*x+c))**(1/2),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}{\sqrt{b \cos \left (d x + c\right )} \cos \left (d x + c\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))/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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