\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [264]

Optimal result

Integrand size = 37, antiderivative size = 169 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8 a^2 (19 A+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (19 A+35 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/7*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+6/35*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+
c)^(3/2)+8/105*a^2*(19*A+35*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/105*a*(19*A+35*C)*sin(d*
x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {4172, 4098, 3894, 3889} \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8 a^2 (19 A+35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (19 A+35 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d \sqrt {\sec (c+d x)}}+\frac {6 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(8*a^2*(19*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(19*A + 35*C)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x]
)/(7*d*Sec[c + d*x]^(5/2)) + (6*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2))

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 3894

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

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -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)}{7 d \sec ^{\frac {5}{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 (2 A+7 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{7 a} \\ & = \frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{35} (19 A+35 C) \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a (19 A+35 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{105} (4 a (19 A+35 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {8 a^2 (19 A+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (19 A+35 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {6 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.50 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (15 A+39 A \sec (c+d x)+(52 A+35 C) \sec ^2(c+d x)+(104 A+175 C) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(2*a^2*(15*A + 39*A*Sec[c + d*x] + (52*A + 35*C)*Sec[c + d*x]^2 + (104*A + 175*C)*Sec[c + d*x]^3)*Sin[c + d*x]
)/(105*d*Sec[c + d*x]^(5/2)*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cos \left (d x +c \right )-1\right ) \left (15 A \cos \left (d x +c \right )^{3}+39 A \cos \left (d x +c \right )^{2}+52 A \cos \left (d x +c \right )+35 C \cos \left (d x +c \right )+104 A +175 C \right ) \csc \left (d x +c \right )}{105 d \sqrt {\sec \left (d x +c \right )}}\)	\(88\)
parts	\(\frac {2 A a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+5 \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(127\)

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*a/d*(a*(1+sec(d*x+c)))^(1/2)*(cos(d*x+c)-1)*(15*A*cos(d*x+c)^3+39*A*cos(d*x+c)^2+52*A*cos(d*x+c)+35*C*c
os(d*x+c)+104*A+175*C)/sec(d*x+c)^(1/2)*csc(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A a \cos \left (d x + c\right )^{4} + 39 \, A a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 35 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (104 \, A + 175 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*A*a*cos(d*x + c)^4 + 39*A*a*cos(d*x + c)^3 + (52*A + 35*C)*a*cos(d*x + c)^2 + (104*A + 175*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. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (145) = 290\).

Time = 0.46 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (735 \, a \cos \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 175 \, a \cos \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, a \cos \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 735 \, a \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 175 \, a \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 63 \, a \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 30 \, a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, a \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 175 \, a \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 735 \, a \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 280 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} C \sqrt {a}}{840 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*(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(si
n(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 175*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*arctan2(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))))*A*sqrt(a) + 280*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a)
)/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 17.67 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (910\,A\,\sin \left (c+d\,x\right )+1400\,C\,\sin \left (c+d\,x\right )+238\,A\,\sin \left (2\,c+2\,d\,x\right )+78\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+140\,C\,\sin \left (2\,c+2\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(7/2),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)*(910*A*sin(c + d*x) + 1400*
C*sin(c + d*x) + 238*A*sin(2*c + 2*d*x) + 78*A*sin(3*c + 3*d*x) + 15*A*sin(4*c + 4*d*x) + 140*C*sin(2*c + 2*d*
x)))/(420*d*(cos(c + d*x) + 1))