\(\int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [155]

Optimal result

Integrand size = 23, antiderivative size = 34 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {b \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 3855} \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {b \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x)) (b \cos (c+d x))^{3/2}}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.77 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {2 \sqrt {\cos \left (d x +c \right ) b}\, b \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(41\)
risch	\(-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.35 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {b^{\frac {3}{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, d}, -\frac {\sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right )}{d}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (30) = 60\).

Time = 0.42 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {{\left (b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} \sqrt {b}}{2 \, d} \]

[In]

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

[Out]

1/2*(b*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*s
in(d*x + c) + 1))*sqrt(b)/d

Giac [F]

\[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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