\(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [245]

Optimal result

Integrand size = 35, antiderivative size = 178 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {64 a^3 (5 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (5 A+7 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/35*a*(5*A+7*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/
d/sec(d*x+c)^(5/2)+64/105*a^3*(5*A+7*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(5*A+7
*B)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {4098, 3894, 3889} \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {64 a^3 (5 A+7 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (5 A+7 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+7 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]

[Out]

(64*a^3*(5*A + 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(5*A + 7*B)*Sq
rt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*a*(5*A + 7*B)*(a + a*Sec[c + d*x])^(3/2)*
Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2
))

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} (5 A+7 B) \int \frac {(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a (5 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{35} (8 a (5 A+7 B)) \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {16 a^2 (5 A+7 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{105} \left (32 a^2 (5 A+7 B)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {64 a^3 (5 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (5 A+7 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 a (5 A+7 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.51 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (15 A+3 (20 A+7 B) \sec (c+d x)+(115 A+98 B) \sec ^2(c+d x)+(230 A+301 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{105 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]

[Out]

(2*a^3*(15*A + 3*(20*A + 7*B)*Sec[c + d*x] + (115*A + 98*B)*Sec[c + d*x]^2 + (230*A + 301*B)*Sec[c + d*x]^3)*S
in[c + d*x])/(105*d*Sec[c + d*x]^(5/2)*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 5.74 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 a^{2} \left (15 A \cos \left (d x +c \right )^{3}+60 A \cos \left (d x +c \right )^{2}+21 B \cos \left (d x +c \right )^{2}+115 A \cos \left (d x +c \right )+98 B \cos \left (d x +c \right )+230 A +301 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(103\)
parts	\(\frac {2 A \,a^{2} \left (3 \cos \left (d x +c \right )^{3}+12 \cos \left (d x +c \right )^{2}+23 \cos \left (d x +c \right )+46\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{21 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+43 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\)	\(147\)

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/105*a^2/d*(15*A*cos(d*x+c)^3+60*A*cos(d*x+c)^2+21*B*cos(d*x+c)^2+115*A*cos(d*x+c)+98*B*cos(d*x+c)+230*A+301*
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 = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (230 \, A + 301 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*A*a^2*cos(d*x + c)^4 + 3*(20*A + 7*B)*a^2*cos(d*x + c)^3 + (115*A + 98*B)*a^2*cos(d*x + c)^2 + (230*
A + 301*B)*a^2*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(c
os(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/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. 385 vs. \(2 (154) = 308\).

Time = 0.53 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.16 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {5 \, \sqrt {2} {\left (315 \, a^{2} \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 ) + 77 \, a^{2} \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 ) + 21 \, a^{2} \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 ) - 315 \, a^{2} \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 ) - 77 \, a^{2} \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 ) - 21 \, a^{2} \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 ) + 6 \, a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, a^{2} \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 ) + 77 \, a^{2} \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 ) + 315 \, a^{2} \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} + 28 \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{840 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*(5*sqrt(2)*(315*a^2*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) +
77*a^2*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) + 21*a^2*cos(2/7*arct
an2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*a
rctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*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))) - 21*a^2*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))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)
)) + 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arctan2(sin(7/2*d*x
+ 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) + 28*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2
*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 16.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.75 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (1960\,A\,\sin \left (c+d\,x\right )+2450\,B\,\sin \left (c+d\,x\right )+490\,A\,\sin \left (2\,c+2\,d\,x\right )+120\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+392\,B\,\sin \left (2\,c+2\,d\,x\right )+42\,B\,\sin \left (3\,c+3\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(7/2),x)

[Out]

(a^2*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(1960*A*sin(c + d*x) + 24
50*B*sin(c + d*x) + 490*A*sin(2*c + 2*d*x) + 120*A*sin(3*c + 3*d*x) + 15*A*sin(4*c + 4*d*x) + 392*B*sin(2*c +
2*d*x) + 42*B*sin(3*c + 3*d*x)))/(420*d*(cos(c + d*x) + 1))