\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) [265]

Optimal result

Integrand size = 37, antiderivative size = 219 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (52 A+63 C) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (136 A+189 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/9*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2/315*a^2*(52*A+63*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)
/(a+a*sec(d*x+c))^(1/2)+2/315*a^2*(136*A+189*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/315*a^2
*(136*A+189*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/21*a*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)
/d/sec(d*x+c)^(5/2)

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4172, 4102, 4100, 3890, 3889} \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (52 A+63 C) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4 a^2 (136 A+189 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[In]

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

[Out]

(2*a^2*(52*A + 63*C)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(136*A + 189*C
)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^2*(136*A + 189*C)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(21*d*Sec[c +
d*x]^(5/2)) + (2*A*(a + a*Sec[c + d*x])^(3/2)*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]

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 (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3 a A}{2}+\frac {1}{2} a (4 A+9 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{9 a} \\ & = \frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (52 A+63 C)+\frac {1}{4} a^2 (40 A+63 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{63 a} \\ & = \frac {2 a^2 (52 A+63 C) \sin (c+d x)}{315 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)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{105} (a (136 A+189 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (52 A+63 C) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+189 C) \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)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {1}{315} (2 a (136 A+189 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^2 (52 A+63 C) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+189 C) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^2 (136 A+189 C) \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)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.47 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (35 A+85 A \sec (c+d x)+3 (34 A+21 C) \sec ^2(c+d x)+(136 A+189 C) \sec ^3(c+d x)+(272 A+378 C) \sec ^4(c+d x)\right ) \sin (c+d x)}{315 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*a^2*(35*A + 85*A*Sec[c + d*x] + 3*(34*A + 21*C)*Sec[c + d*x]^2 + (136*A + 189*C)*Sec[c + d*x]^3 + (272*A +
378*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 = 1.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.51

method	result	size
default	\(\frac {2 a \left (35 A \cos \left (d x +c \right )^{4}+85 A \cos \left (d x +c \right )^{3}+102 A \cos \left (d x +c \right )^{2}+63 C \cos \left (d x +c \right )^{2}+136 A \cos \left (d x +c \right )+189 C \cos \left (d x +c \right )+272 A +378 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(112\)
parts	\(\frac {2 A a \left (35 \cos \left (d x +c \right )^{4}+85 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+136 \cos \left (d x +c \right )+272\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+3 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\)	\(151\)

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/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+102*A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+136*A*cos(d*x+c)+189*C*cos
(d*x+c)+272*A+378*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 = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (35 \, A a \cos \left (d x + c\right )^{5} + 85 \, A a \cos \left (d x + c\right )^{4} + 3 \, {\left (34 \, A + 21 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (136 \, A + 189 \, C\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (136 \, A + 189 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

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

[Out]

2/315*(35*A*a*cos(d*x + c)^5 + 85*A*a*cos(d*x + c)^4 + 3*(34*A + 21*C)*a*cos(d*x + c)^3 + (136*A + 189*C)*a*co
s(d*x + c)^2 + 2*(136*A + 189*C)*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. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**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. 607 vs. \(2 (189) = 378\).

Time = 0.49 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.77 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/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) + 252*sqrt(2)*(20*a*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*a*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) - 20*a*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*a*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))) + 2*a*s
in(5/2*d*x + 5/2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 20*a*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]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 18.90 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.65 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (4830\,A\,\sin \left (c+d\,x\right )+6300\,C\,\sin \left (c+d\,x\right )+1428\,A\,\sin \left (2\,c+2\,d\,x\right )+513\,A\,\sin \left (3\,c+3\,d\,x\right )+170\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+1512\,C\,\sin \left (2\,c+2\,d\,x\right )+252\,C\,\sin \left (3\,c+3\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(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) + 6300
*C*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) + 1512*C*sin(2*c + 2*d*x) + 252*C*sin(3*c + 3*d*x)))/(2520*d*(cos(c + d*x) + 1))