\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [140]

Optimal result

Integrand size = 33, antiderivative size = 154 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (19 A+20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-4 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \]

[Out]

1/4*a^(5/2)*(19*A+20*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/2*a*A*cos(d*x+c)*(a+a*sec(d*x+c)
)^(3/2)*sin(d*x+c)/d+1/4*a^3*(9*A-4*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/2*a^2*(A-4*B)*sin(d*x+c)*(a+a*sec
(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4102, 4103, 4100, 3859, 209} \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (19 A+20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (A-4 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{2 d} \]

[In]

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

[Out]

(a^(5/2)*(19*A + 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(9*A - 4*B)*Sin[c
+ d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 4*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + (a*A*C
os[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*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]

Rule 4103

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m +
n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d
*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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 (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (7 A+4 B)-\frac {1}{2} a (A-4 B) \sec (c+d x)\right ) \, dx \\ & = -\frac {a^2 (A-4 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (9 A-4 B)+\frac {1}{4} a^2 (5 A+12 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-4 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac {1}{8} \left (a^2 (19 A+20 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-4 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}-\frac {\left (a^3 (19 A+20 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 )}{4 d} \\ & = \frac {a^{5/2} (19 A+20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (9 A-4 B) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-4 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.56 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=-\frac {a^2 \left (-\left ((7 A+20 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )\right )+(3 A-8 B+(A-4 B) \cos (c+d x)+3 A \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)}-32 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

-1/4*(a^2*(-((7*A + 20*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]) + (3*A - 8*B + (A - 4*B)*Cos[c + d*x] + 3*A*Cos[2*(
c + d*x)])*Sqrt[1 - Sec[c + d*x]] - 32*A*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x
]])*Sqrt[a*(1 + Sec[c + d*x])]*Sin[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. \(343\) vs. \(2(134)=268\).

Time = 98.45 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.23

method	result	size
default	\(\frac {a^{2} \left (19 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+20 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+19 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+11 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+20 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+4 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \left (\cos \left (d x +c \right )+1\right )}\)	\(344\)

[In]

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

[Out]

1/4*a^2/d*(19*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)+2*A*cos(d*x+c)^2*sin(d*x+c)+20*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)+19*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(s
in(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+11*A*cos(d*x+c)*sin(d*x+c)+20*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)+4*B*cos(d*x+c)*sin(d*x
+c)+8*B*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.26 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {{\left ({\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (19 \, A + 20 \, B\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (19 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (19 \, A + 20 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

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

[Out]

[1/8*(((19*A + 20*B)*a^2*cos(d*x + c) + (19*A + 20*B)*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*(2*
A*a^2*cos(d*x + c)^2 + (11*A + 4*B)*a^2*cos(d*x + c) + 8*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*
x + c))/(d*cos(d*x + c) + d), -1/4*(((19*A + 20*B)*a^2*cos(d*x + c) + (19*A + 20*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))) - (2*A*a^2*cos(d*x + c)^2 + (11*A + 4*B
)*a^2*cos(d*x + c) + 8*B*a^2)*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. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[In]

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

[Out]

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