\(\int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx\) [877]

Optimal result

Integrand size = 33, antiderivative size = 74 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {18, 2827, 3852, 8, 3855} \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {A \sin (c+d x)}{b d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \text {arctanh}(\sin (c+d x))}{b d \sqrt {b \cos (c+d x)}} \]

[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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2827

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 3852

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 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 ^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 \text {arctanh}(\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 ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{b d \sqrt {b \cos (c+d x)}} \\ & = \frac {B \text {arctanh}(\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] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos (c+d x)} (B \text {arctanh}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x))}{d (b \cos (c+d x))^{3/2}} \]

[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))

Maple [A] (verified)

Time = 5.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {-2 B \cos \left (d x +c \right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+A \sin \left (d x +c \right )}{b d \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\)	\(60\)
parts	\(\frac {A \sin \left (d x +c \right )}{b d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}-\frac {2 B \left (\sqrt {\cos }\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}\, b}\)	\(77\)
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{b \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(118\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.85 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\left [\frac {B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )^{2}}, -\frac {B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d \cos \left (d x + c\right )^{2}}\right ] \]

[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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (66) = 132\).

Time = 0.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.80 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}} + \frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {3}{2}}}}{2 \, d} \]

[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

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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