\(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [876]

Optimal result

Integrand size = 33, antiderivative size = 66 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {B x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {17, 2814, 3855} \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {A \sqrt {\cos (c+d x)} \text {arctanh}(\sin (c+d x))}{b d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(b*Sqrt[b*Cos[c + d*x]]) + (A*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(b*d*Sqrt[b*C
os[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*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \sec (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}} \\ & = \frac {B x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {\left (A \sqrt {\cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}} \\ & = \frac {B x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {(B d x+A \text {arctanh}(\sin (c+d x))) \cos ^{\frac {3}{2}}(c+d x)}{d (b \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 5.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {\left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-B \left (d x +c \right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}\)	\(55\)
parts	\(-\frac {2 A \left (\sqrt {\cos }\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}\, b}+\frac {B \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d b \sqrt {\cos \left (d x +c \right ) b}}\)	\(76\)
risch	\(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(105\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.26 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\left [-\frac {2 \, A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) + B \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, b^{2} d}, \frac {2 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + A \sqrt {b} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, b^{2} d}\right ] \]

[In]

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

Sympy [F]

\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {A {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {3}{2}}} + \frac {4 \, B \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {3}{2}}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),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^(3/2) + 4*B*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^(3/2))/d

Giac [F]

\[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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