\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [141]

Optimal result

Integrand size = 33, antiderivative size = 164 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (25 A+38 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \]

[Out]

1/8*a^(5/2)*(25*A+38*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/3*a*A*cos(d*x+c)^2*(a+a*sec(d*x+
c))^(3/2)*sin(d*x+c)/d+1/24*a^3*(49*A+54*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a^2*(3*A+2*B)*cos(d*x+c)*s
in(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4102, 4100, 3859, 209} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (25 A+38 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[In]

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

[Out]

(a^(5/2)*(25*A + 38*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(49*A + 54*B)*Sin
[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(3*A + 2*B)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*
x])/(4*d) + (a*A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*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 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]

Rule 4102

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

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (A+6 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (49 A+54 B)+\frac {1}{4} a^2 (13 A+30 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{16} \left (a^2 (25 A+38 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {\left (a^3 (25 A+38 B)\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 {a^{5/2} (25 A+38 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a A \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \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 = 0.75 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.90 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {a^2 \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left (-165 A \sqrt {1-\sec (c+d x)} \sin (c+d x)+18 B \sqrt {1-\sec (c+d x)} \sin (c+d x)+8 A \cos ^2(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)-31 A \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))+54 B \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))-165 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-126 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-576 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)-192 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)\right )}{72 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

-1/72*(a^2*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(-165*A*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 18*B*Sqrt[1 -
Sec[c + d*x]]*Sin[c + d*x] + 8*A*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] - 31*A*Sqrt[1 - Sec[c + d
*x]]*Sin[2*(c + d*x)] + 54*B*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] - 165*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*T
an[c + d*x] - 126*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 576*B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Se
c[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x] - 192*A*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[
1 - Sec[c + d*x]]*Tan[c + d*x]))/(d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(368\) vs. \(2(144)=288\).

Time = 258.82 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.25

method	result	size
default	\(\frac {a^{2} \left (8 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+75 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+34 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+114 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+12 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+75 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+114 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+66 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{24 d \left (\cos \left (d x +c \right )+1\right )}\)	\(369\)

[In]

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

[Out]

1/24*a^2/d*(8*A*sin(d*x+c)*cos(d*x+c)^3+75*A*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)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+34*A*cos(d*x+c)^2*sin(d*x+c)+114*B*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)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+12*B*sin(d*x+c)*co
s(d*x+c)^2+75*A*(-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*A*cos(d*x+c)*sin(d*x+c)+114*B*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)/(cos(d*x+c)+1))^(1/2)+66*B*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 = 0.32 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.32 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

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

[Out]

[1/48*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A + 38*B)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sq
rt((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*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B)*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*(3*((25*A + 38*B)*a^2*cos(d*x + c) + (25*A
+ 38*B)*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*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B)*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. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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