\(\int \frac {\sqrt {\cos (c+d x)} (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [127]

Optimal result

Integrand size = 35, antiderivative size = 74 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 3093, 3855} \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {A \sqrt {\cos (c+d x)} \text {arctanh}(\sin (c+d x))}{b d \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}} \]

[In]

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

[Out]

(A*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(b*d*Sqrt[b*Cos[c + d*x]]) + (C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/
(b*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 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}} \\ & = \frac {C \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \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 {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) (A \text {arctanh}(\sin (c+d x))+C \sin (c+d x))}{d (b \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.38 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-\sin \left (d x +c \right ) C \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}\)	\(56\)
parts	\(-\frac {2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{d b \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}\)	\(77\)
risch	\(-\frac {i \left (\sqrt {\cos }\left (d x +c \right )\right ) C \,{\mathrm e}^{i \left (d x +c \right )}}{2 b \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {i \left (\sqrt {\cos }\left (d x +c \right )\right ) C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b \sqrt {\cos \left (d x +c \right ) b}\, d}\)	\(155\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.80 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\left [\frac {A \sqrt {b} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )}, -\frac {A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right ) - \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {A {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {3}{2}}} + \frac {2 \, C \sin \left (d x + c\right )}{b^{\frac {3}{2}}}}{2 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(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) + 2*C*sin(d*x + c)/b^(3/2))/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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