\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) [257]

Optimal result

Integrand size = 37, antiderivative size = 213 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (16 A+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/63*a*A*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2/105*a*(16*A+21*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2
)/(a+a*sec(d*x+c))^(1/2)+8/315*a*(16*A+21*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/315*a*(16
*A+21*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/9*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*
x+c)^(7/2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {4172, 4100, 3890, 3889} \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a (16 A+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]

[In]

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

[Out]

(2*a*A*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 21*C)*Sin[c + d*x])/(10
5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*
Sqrt[a + a*Sec[c + d*x]]) + (16*a*(16*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x
]]) + (2*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 4172

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a A}{2}+\frac {3}{2} a (2 A+3 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{21} (16 A+21 C) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 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 (16 A+21 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 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 (16 A+21 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a A \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a (16 A+21 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 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 = 0.97 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a \left (35 A+40 A \sec (c+d x)+(48 A+63 C) \sec ^2(c+d x)+(64 A+84 C) \sec ^3(c+d x)+8 (16 A+21 C) \sec ^4(c+d x)\right ) \sin (c+d x)}{315 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*a*(35*A + 40*A*Sec[c + d*x] + (48*A + 63*C)*Sec[c + d*x]^2 + (64*A + 84*C)*Sec[c + d*x]^3 + 8*(16*A + 21*C)
*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 = 0.91 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {2 \left (35 A \cos \left (d x +c \right )^{4}+40 A \cos \left (d x +c \right )^{3}+48 A \cos \left (d x +c \right )^{2}+63 C \cos \left (d x +c \right )^{2}+64 A \cos \left (d x +c \right )+84 C \cos \left (d x +c \right )+128 A +168 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(111\)
parts	\(\frac {2 A \left (35 \cos \left (d x +c \right )^{4}+40 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+64 \cos \left (d x +c \right )+128\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\)	\(151\)

[In]

int((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/315/d*(35*A*cos(d*x+c)^4+40*A*cos(d*x+c)^3+48*A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+64*A*cos(d*x+c)+84*C*cos(d*x+
c)+128*A+168*C)*(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 = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (35 \, A \cos \left (d x + c\right )^{5} + 40 \, A \cos \left (d x + c\right )^{4} + 3 \, {\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/315*(35*A*cos(d*x + c)^5 + 40*A*cos(d*x + c)^4 + 3*(16*A + 21*C)*cos(d*x + c)^3 + 4*(16*A + 21*C)*cos(d*x +
c)^2 + 8*(16*A + 21*C)*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. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((A+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/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. 587 vs. \(2 (183) = 366\).

Time = 0.50 (sec) , antiderivative size = 587, normalized size of antiderivative = 2.76 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(1890*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) + 420*
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) + 252*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) + 45*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) - 1890*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))) - 420*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))) - 252*
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))) - 45*cos(9/2*d*x + 9/2*c)*si
n(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*sin(9/2*d*x + 9/2*c) + 45*sin(7/9*arctan2(sin(
9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) +
420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 1890*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c),
cos(9/2*d*x + 9/2*c))))*A*sqrt(a) + 84*sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))
)*sin(5/2*d*x + 5/2*c) + 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) -
30*cos(5/2*d*x + 5/2*c)*sin(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)
*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin
(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*C
*sqrt(a))/d

Giac [F]

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

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 18.67 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (2310\,A\,\sin \left (c+d\,x\right )+2940\,C\,\sin \left (c+d\,x\right )+672\,A\,\sin \left (2\,c+2\,d\,x\right )+297\,A\,\sin \left (3\,c+3\,d\,x\right )+80\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+672\,C\,\sin \left (2\,c+2\,d\,x\right )+252\,C\,\sin \left (3\,c+3\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

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

[Out]

(cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(2310*A*sin(c + d*x) + 2940*C
*sin(c + d*x) + 672*A*sin(2*c + 2*d*x) + 297*A*sin(3*c + 3*d*x) + 80*A*sin(4*c + 4*d*x) + 35*A*sin(5*c + 5*d*x
) + 672*C*sin(2*c + 2*d*x) + 252*C*sin(3*c + 3*d*x)))/(2520*d*(cos(c + d*x) + 1))