3.9.83 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [883]

Optimal. Leaf size=66 \[ \frac {B x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}}+\frac {A \tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

B*x*cos(d*x+c)^(1/2)/b^2/(b*cos(d*x+c))^(1/2)+A*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 = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {17, 2814, 3855} \begin {gather*} \frac {A \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b^2 d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(b^2*Sqrt[b*Cos[c + d*x]]) + (A*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(b^2*d*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 2814

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 3855

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

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Mathematica [A]
time = 0.04, size = 43, normalized size = 0.65 \begin {gather*} \frac {\left (B d x+A \tanh ^{-1}(\sin (c+d x))\right ) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 54, normalized size = 0.82

method	result	size
default	\(-\frac {\left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (2 A \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-B \left (d x +c \right )\right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(54\)
risch	\(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} \sqrt {b \cos \left (d x +c \right )}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.57, size = 92, normalized size = 1.39 \begin {gather*} \frac {\frac {A {\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 \, B \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {5}{2}}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(A*(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*si
n(d*x + c) + 1))/b^(5/2) + 4*B*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^(5/2))/d

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Fricas [A]
time = 0.45, size = 215, normalized size = 3.26 \begin {gather*} \left [-\frac {2 \, 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 ) + B \sqrt {-b} \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 \, b^{3} d}, \frac {2 \, B \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 ) + A \sqrt {b} \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 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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