3.256 \(\int \frac {\sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=168 \[ \frac {4 a (24 A+35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a A \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]

[Out]

2/35*a*A*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/105*a*(24*A+35*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2
)/(a+a*sec(d*x+c))^(1/2)+4/105*a*(24*A+35*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/7*A*sin(d*
x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)

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Rubi [A]  time = 0.39, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {4087, 4015, 3805, 3804} \[ \frac {4 a (24 A+35 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a A \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(24*A + 35*C)*Sin[c + d*x])/(10
5*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*(24*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*
Sqrt[a + a*Sec[c + d*x]]) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

Rule 3804

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 3805

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 4015

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 4087

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*} \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a A}{2}+\frac {1}{2} a (4 A+7 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 a A \sin (c+d x)}{35 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)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{35} (24 A+35 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)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (24 A+35 C) \sin (c+d x)}{105 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)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{105} (2 (24 A+35 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (24 A+35 C) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a (24 A+35 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 84, normalized size = 0.50 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} ((141 A+140 C) \cos (c+d x)+36 A \cos (2 (c+d x))+15 A \cos (3 (c+d x))+228 A+280 C)}{210 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((228*A + 280*C + (141*A + 140*C)*Cos[c + d*x] + 36*A*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[a*(1 + Se
c[c + d*x])]*Tan[(c + d*x)/2])/(210*d*Sqrt[Sec[c + d*x]])

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fricas [A]  time = 0.44, size = 103, normalized size = 0.61 \[ \frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} + 18 \, A \cos \left (d x + c\right )^{3} + {\left (24 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (24 \, A + 35 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*A*cos(d*x + c)^4 + 18*A*cos(d*x + c)^3 + (24*A + 35*C)*cos(d*x + c)^2 + 2*(24*A + 35*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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 3.70, size = 107, normalized size = 0.64 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (15 A \left (\cos ^{3}\left (d x +c \right )\right )+18 A \left (\cos ^{2}\left (d x +c \right )\right )+24 A \cos \left (d x +c \right )+35 C \cos \left (d x +c \right )+48 A +70 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{105 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(7/2),x)

[Out]

-2/105/d*(-1+cos(d*x+c))*(15*A*cos(d*x+c)^3+18*A*cos(d*x+c)^2+24*A*cos(d*x+c)+35*C*cos(d*x+c)+48*A+70*C)*(a*(1
+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)

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maxima [B]  time = 0.69, size = 408, normalized size = 2.43 \[ \frac {3 \, \sqrt {2} {\left (105 \, \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 ) + 35 \, \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 ) + 7 \, \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 ) - 105 \, \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 ) - 35 \, \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 ) - 7 \, \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 ) + 10 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \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 ) + 35 \, \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 ) + 105 \, \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} + 140 \, \sqrt {2} {\left (3 \, \cos \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \sin \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 2 \, \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} C \sqrt {a}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/840*(3*sqrt(2)*(105*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) + 35*c
os(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) + 7*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) - 105*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))) - 35*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))) - 7*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))) + 10*sin(7/2
*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*sin(3/7*arctan2(sin(7/2*d*
x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(
a) + 140*sqrt(2)*(3*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(
3/2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*sin
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*C*sqrt(a))/d

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mupad [B]  time = 4.54, size = 118, normalized size = 0.70 \[ \frac {\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (420\,A\,\sin \left (c+d\,x\right )+560\,C\,\sin \left (c+d\,x\right )+126\,A\,\sin \left (2\,c+2\,d\,x\right )+36\,A\,\sin \left (3\,c+3\,d\,x\right )+15\,A\,\sin \left (4\,c+4\,d\,x\right )+140\,C\,\sin \left (2\,c+2\,d\,x\right )\right )}{420\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/(1/cos(c + d*x))^(7/2),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)*(420*A*sin(c + d*x) + 560*C*s
in(c + d*x) + 126*A*sin(2*c + 2*d*x) + 36*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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/sec(d*x+c)**(7/2),x)

[Out]

Timed out

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