3.236 \(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=131 \[ \frac {8 a^2 (3 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (3 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/5*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/15*a^2*(3*A+5*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a
+a*sec(d*x+c))^(1/2)+2/15*a*(3*A+5*B)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {4013, 3809, 3804} \[ \frac {8 a^2 (3 A+5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (3 A+5 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(8*a^2*(3*A + 5*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 5*B)*Sqrt[a
+ a*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d
*Sec[c + d*x]^(3/2))

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 3809

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*Co
t[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*C
sc[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 4013

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*} \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} (3 A+5 B) \int \frac {(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{15} (4 a (3 A+5 B)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {8 a^2 (3 A+5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+5 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 73, normalized size = 0.56 \[ \frac {a^2 \sin (c+d x) \sqrt {\sec (c+d x)} (2 (9 A+5 B) \cos (c+d x)+3 A \cos (2 (c+d x))+39 A+50 B)}{15 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]

[Out]

(a^2*(39*A + 50*B + 2*(9*A + 5*B)*Cos[c + d*x] + 3*A*Cos[2*(c + d*x)])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*
Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.42, size = 94, normalized size = 0.72 \[ \frac {2 \, {\left (3 \, A a \cos \left (d x + c\right )^{3} + {\left (9 \, A + 5 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (18 \, A + 25 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*A*a*cos(d*x + c)^3 + (9*A + 5*B)*a*cos(d*x + c)^2 + (18*A + 25*B)*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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(5/2), x)

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maple [A]  time = 2.62, size = 97, normalized size = 0.74 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3 A \left (\cos ^{2}\left (d x +c \right )\right )+9 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+18 A +25 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} a}{15 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(3*A*cos(d*x+c)^2+9*A*cos(d*x+c)+5*B*cos(d*x+c)+18*A+25*B)*(a*(1+cos(d*x+c))/cos(d*x+c
))^(1/2)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)*a

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maxima [B]  time = 1.50, size = 250, normalized size = 1.91 \[ \frac {3 \, \sqrt {2} {\left (20 \, a \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 20 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 5 \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 2 \, a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 20 \, a \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 20 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/60*(3*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*sin(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))))*A*sqrt(a) + 20*(sq
rt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

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mupad [B]  time = 3.40, size = 107, normalized size = 0.82 \[ \frac {a\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (75\,A\,\sin \left (c+d\,x\right )+100\,B\,\sin \left (c+d\,x\right )+18\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )\right )}{30\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2))/(1/cos(c + d*x))^(5/2),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)*(75*A*sin(c + d*x) + 100*B*
sin(c + d*x) + 18*A*sin(2*c + 2*d*x) + 3*A*sin(3*c + 3*d*x) + 10*B*sin(2*c + 2*d*x)))/(30*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+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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