\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) [238]

Optimal result

Integrand size = 35, antiderivative size = 228 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (34 A+39 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/63*a^2*(10*A+9*B)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2/105*a^2*(34*A+39*B)*sin(d*x+c)/d/se
c(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+8/315*a^2*(34*A+39*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/
2)+16/315*a^2*(34*A+39*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/9*a*A*sin(d*x+c)*(a+a*sec(d*x
+c))^(1/2)/d/sec(d*x+c)^(7/2)

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4102, 4100, 3890, 3889} \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (34 A+39 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x]

[Out]

(2*a^2*(10*A + 9*B)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(34*A + 39*B)*Si
n[c + d*x])/(105*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(34*A + 39*B)*Sin[c + d*x])/(315*d*Sq
rt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(34*A + 39*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqr
t[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, 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 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 {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2}{9} \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (10 A+9 B)+\frac {3}{2} a (2 A+3 B) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} (a (34 A+39 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{105} (4 a (34 A+39 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{315} (8 a (34 A+39 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^2 (10 A+9 B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (34 A+39 B) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (34 A+39 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (34 A+39 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.48 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (35 A+5 (17 A+9 B) \sec (c+d x)+3 (34 A+39 B) \sec ^2(c+d x)+4 (34 A+39 B) \sec ^3(c+d x)+8 (34 A+39 B) \sec ^4(c+d x)\right ) \sin (c+d x)}{315 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(9/2),x]

[Out]

(2*a^2*(35*A + 5*(17*A + 9*B)*Sec[c + d*x] + 3*(34*A + 39*B)*Sec[c + d*x]^2 + 4*(34*A + 39*B)*Sec[c + d*x]^3 +
8*(34*A + 39*B)*Sec[c + d*x]^4)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(7/2)*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 5.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.54

method	result	size
default	\(\frac {2 a \left (35 A \cos \left (d x +c \right )^{4}+85 A \cos \left (d x +c \right )^{3}+45 B \cos \left (d x +c \right )^{3}+102 A \cos \left (d x +c \right )^{2}+117 B \cos \left (d x +c \right )^{2}+136 A \cos \left (d x +c \right )+156 B \cos \left (d x +c \right )+272 A +312 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(123\)
parts	\(\frac {2 A a \left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(158\)

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/315*a/d*(35*A*cos(d*x+c)^4+85*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+102*A*cos(d*x+c)^2+117*B*cos(d*x+c)^2+136*A*c
os(d*x+c)+156*B*cos(d*x+c)+272*A+312*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.58 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (35 \, A a \cos \left (d x + c\right )^{5} + 5 \, {\left (17 \, A + 9 \, B\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )^{3} + 4 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (34 \, A + 39 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/315*(35*A*a*cos(d*x + c)^5 + 5*(17*A + 9*B)*a*cos(d*x + c)^4 + 3*(34*A + 39*B)*a*cos(d*x + c)^3 + 4*(34*A +
39*B)*a*cos(d*x + c)^2 + 8*(34*A + 39*B)*a*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)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (198) = 396\).

Time = 0.54 (sec) , antiderivative size = 700, normalized size of antiderivative = 3.07 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(3780*a*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 10
50*a*cos(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 378*a*cos(4/9*arctan2
(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 135*a*cos(2/9*arctan2(sin(9/2*d*x + 9/2*c
), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 3780*a*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x + 9/2
*c), cos(9/2*d*x + 9/2*c))) - 1050*a*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x +
9/2*c))) - 378*a*cos(9/2*d*x + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 135*a*cos
(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*a*sin(9/2*d*x + 9/2*c) + 1
35*a*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c)
, cos(9/2*d*x + 9/2*c))) + 1050*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 3780*a*sin(1/
9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) + 6*sqrt(2)*(735*a*cos(6/7*arctan2(sin(7/2*d
*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 175*a*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 63*a*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7
/2*d*x + 7/2*c) - 735*a*cos(7/2*d*x + 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 17
5*a*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 63*a*cos(7/2*d*x + 7/2
*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 30*a*sin(7/2*d*x + 7/2*c) + 63*a*sin(5/7*ar
ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x +
7/2*c))) + 735*a*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*B*sqrt(a))/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 17.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (4830\,A\,\sin \left (c+d\,x\right )+5460\,B\,\sin \left (c+d\,x\right )+1428\,A\,\sin \left (2\,c+2\,d\,x\right )+513\,A\,\sin \left (3\,c+3\,d\,x\right )+170\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+1428\,B\,\sin \left (2\,c+2\,d\,x\right )+468\,B\,\sin \left (3\,c+3\,d\,x\right )+90\,B\,\sin \left (4\,c+4\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(9/2),x)

[Out]

(a*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(4830*A*sin(c + d*x) + 5460
*B*sin(c + d*x) + 1428*A*sin(2*c + 2*d*x) + 513*A*sin(3*c + 3*d*x) + 170*A*sin(4*c + 4*d*x) + 35*A*sin(5*c + 5
*d*x) + 1428*B*sin(2*c + 2*d*x) + 468*B*sin(3*c + 3*d*x) + 90*B*sin(4*c + 4*d*x)))/(2520*d*(cos(c + d*x) + 1))