\(\int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [846]

Optimal result

Integrand size = 33, antiderivative size = 60 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {B x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, 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} \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {B x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(B d x+A \text {arctanh}(\sin (c+d x))) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-B \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(52\)
parts	\(-\frac {2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(70\)
risch	\(\frac {B x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {\sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {\sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\)	\(96\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (52) = 104\).

Time = 0.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.50 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [-\frac {2 \, A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) - B \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, d}, \frac {2 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + A \sqrt {b} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, d}\right ] \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.53 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {A \sqrt {b} {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} + 4 \, B \sqrt {b} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(A*sqrt(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)) + 4*B*sqrt(b)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)))/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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