\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [114]

Optimal result

Integrand size = 23, antiderivative size = 144 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {25 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {25 a^3 \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

25/8*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+25/8*a^3*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+
13/12*a^3*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a^2*cos(d*x+c)^2*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2
)/d

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3898, 4100, 3890, 3859, 209} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {25 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {25 a^3 \sin (c+d x)}{8 d \sqrt {a \sec (c+d x)+a}}+\frac {13 a^3 \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d} \]

[In]

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

[Out]

(25*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (25*a^3*Sin[c + d*x])/(8*d*Sqrt[a
+ a*Sec[c + d*x]]) + (13*a^3*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]^2
*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3890

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e
+ f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[a*((2*n + 1)/(2*b*d*n)), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 3898

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b^2*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 4100

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} a \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {13 a}{2}+\frac {9}{2} a \sec (c+d x)\right ) \, dx \\ & = \frac {13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{8} \left (25 a^2\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {25 a^3 \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{16} \left (25 a^2\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {25 a^3 \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}-\frac {\left (25 a^3\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d} \\ & = \frac {25 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {25 a^3 \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {13 a^3 \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.84 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {a^2 \left (165 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+(31+159 \cos (c+d x)+31 \cos (2 (c+d x))-2 \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)}+192 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{72 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

(a^2*(165*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + (31 + 159*Cos[c + d*x] + 31*Cos[2*(c + d*x)] - 2*Cos[3*(c + d*x)])
*Sqrt[1 - Sec[c + d*x]] + 192*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*
(1 + Sec[c + d*x])]*Sin[c + d*x])/(72*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

Maple [A] (verified)

Time = 157.07 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {a^{2} \left (8 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+75 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+34 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+75 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+75 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (\cos \left (d x +c \right )+1\right )}\)	\(204\)

[In]

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

[Out]

1/24/d*a^2*(8*cos(d*x+c)^3*sin(d*x+c)+75*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/
(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+34*cos(d*x+c)^2*sin(d*x+c)+75*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+75*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*
x+c)))^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.22 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {75 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{3} + 34 \, a^{2} \cos \left (d x + c\right )^{2} + 75 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {75 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, a^{2} \cos \left (d x + c\right )^{3} + 34 \, a^{2} \cos \left (d x + c\right )^{2} + 75 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

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

[Out]

[1/48*(75*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(8*a^2*cos(d*x + c)^3 + 34*
a^2*cos(d*x + c)^2 + 75*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c
) + d), -1/24*(75*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)
/(sqrt(a)*sin(d*x + c))) - (8*a^2*cos(d*x + c)^3 + 34*a^2*cos(d*x + c)^2 + 75*a^2*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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