\(\int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [210]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {10 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

2/3*a^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+10/3*a^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d
*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2841, 21, 2850} \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {10 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(2*a^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (10*a^2*Sin[c + d*x])/(3*d*Sqrt[Cos[c
+ d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {1}{3} (2 a) \int \frac {-\frac {5 a}{2}-\frac {5}{2} a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx \\ & = \frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {1}{3} (5 a) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {10 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a \sqrt {a (1+\cos (c+d x))} (1+5 \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(1 + 5*Cos[c + d*x])*Tan[(c + d*x)/2])/(3*d*Cos[c + d*x]^(3/2))

Maple [A] (verified)

Time = 5.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{3 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(53\)

[In]

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

[Out]

2/3/d*sin(d*x+c)*(5*cos(d*x+c)+1)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/cos(d*x+c)^(3/2)*a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, {\left (5 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

2/3*(5*a*cos(d*x + c) + a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(
d*x + c)^2)

Sympy [F]

\[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral((a*(cos(c + d*x) + 1))**(3/2)/cos(c + d*x)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.54 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \, {\left (\frac {3 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

4/3*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
+ 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin
(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 15.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (5\,\sin \left (c+d\,x\right )+2\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (3\,\cos \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )+2\right )} \]

[In]

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

[Out]

(2*a*(a*(cos(c + d*x) + 1))^(1/2)*(5*sin(c + d*x) + 2*sin(2*c + 2*d*x) + 5*sin(3*c + 3*d*x)))/(3*d*cos(c + d*x
)^(1/2)*(3*cos(c + d*x) + 2*cos(2*c + 2*d*x) + cos(3*c + 3*d*x) + 2))