\(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [861]

Optimal result

Integrand size = 33, antiderivative size = 65 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {A b^2 x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {17, 2717} \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {A b^2 x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

(A*b^2*x*Sqrt[b*Cos[c + d*x]])/Sqrt[Cos[c + d*x]] + (b^2*B*Sqrt[b*Cos[c + d*x]]*Sin[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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int (A+B \cos (c+d x)) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {A b^2 x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {\left (b^2 B \sqrt {b \cos (c+d x)}\right ) \int \cos (c+d x) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {A b^2 x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (A (c+d x)+B \sin (c+d x))}{d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(A*(c + d*x) + B*Sin[c + d*x]))/(d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 4.91 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (A \left (d x +c \right )+B \sin \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(42\)
risch	\(\frac {A \,b^{2} x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(58\)
parts	\(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.92 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {A \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} B b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, \frac {A b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + \sqrt {b \cos \left (d x + c\right )} B b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[1/2*(A*sqrt(-b)*b^2*cos(d*x + c)*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) + 2*sqrt(b*cos(d*x + c))*B*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), (A*b^(5/2)
*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + sqrt(b*cos(d*x + c))*B*
b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, A b^{\frac {5}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + B b^{\frac {5}{2}} \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

(2*A*b^(5/2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + B*b^(5/2)*sin(d*x + c))/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.58 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+A\,d\,x\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

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