3.620 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=344 \[ -\frac {\left (-3 a^2 B+4 a A b-4 b^2 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {(4 A b-3 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{4 b^2 d \sqrt {\cos (c+d x)}}-\frac {(4 A b-3 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {(4 A b-a B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {B \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{2 b d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/4*(4*A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))
^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/b/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-1/4*(4*A*a*b-3*B*a^2-4*B*b^
2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((
b+a*cos(d*x+c))/(a+b))^(1/2)/b^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+1/2*B*sin(d*x+c)*(a+b*sec(d*x+c))^(
1/2)/b/d/cos(d*x+c)^(3/2)+1/4*(4*A*b-3*B*a)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/b^2/d/cos(d*x+c)^(1/2)-1/4*(4*A*
b-3*B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))
*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/b^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

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Rubi [A]  time = 1.28, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2955, 4033, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac {\left (-3 a^2 B+4 a A b-4 b^2 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {(4 A b-3 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{4 b^2 d \sqrt {\cos (c+d x)}}-\frac {(4 A b-3 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {(4 A b-a B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {B \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{2 b d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(4*b*d*Sqrt[Cos[c + d
*x]]*Sqrt[a + b*Sec[c + d*x]]) - ((4*a*A*b - 3*a^2*B - 4*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[
2, (c + d*x)/2, (2*a)/(a + b)])/(4*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - ((4*A*b - 3*a*B)*Sqrt[
Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(4*b^2*d*Sqrt[(b + a*Cos[c + d*x
])/(a + b)]) + (B*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(2*b*d*Cos[c + d*x]^(3/2)) + ((4*A*b - 3*a*B)*Sqrt[a
+ b*Sec[c + d*x]]*Sin[c + d*x])/(4*b^2*d*Sqrt[Cos[c + d*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4033

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2))/(
b*f*(m + n)), x] + Dist[d^2/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2)
+ B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] &&  !IGtQ[m, 1]

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4102

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

Rule 4108

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {a B}{2}+b B \sec (c+d x)+\frac {1}{2} (4 A b-3 a B) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 b}\\ &=\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a (4 A b-3 a B)+\frac {1}{2} a b B \sec (c+d x)-\frac {1}{4} \left (4 a A b-3 a^2 B-4 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{2 b^2}\\ &=\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a (4 A b-3 a B)+\frac {1}{2} a b B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{2 b^2}+\frac {\left (\left (-4 a A b+3 a^2 B+4 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 b^2}\\ &=\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {\left ((4 A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 b}+\frac {\left ((-4 A b+3 a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{8 b^2}+\frac {\left (\left (-4 a A b+3 a^2 B+4 b^2 B\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{8 b^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ &=\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {\left ((4 A b-a B) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (-4 a A b+3 a^2 B+4 b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{8 b^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left ((-4 A b+3 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{8 b^2 \sqrt {b+a \cos (c+d x)}}\\ &=-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}+\frac {\left ((4 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left ((-4 A b+3 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{8 b^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {(4 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 a A b-3 a^2 B-4 b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{4 b^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {(4 A b-3 a B) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{4 b^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {B \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{2 b d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(4 A b-3 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{4 b^2 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 32.92, size = 77909, normalized size = 226.48 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [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+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.36, size = 1569, normalized size = 4.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(-1+cos(d*x+c))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))*(4*A*EllipticE((-1+cos(d*x+c))*((a-b
)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d
*x+c)^2*a*b-4*A*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*b^2+8*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*E
llipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x
+c)^2*a*b-8*A*cos(d*x+c)^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b
))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a*b-4*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b*(1/(1+cos(d*
x+c)))^(1/2)-3*B*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*a^2+3*B*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/
sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*a*b-6*B
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)
/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)^2*a^2-8*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+
c)^2*b^2+6*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*cos(d*x
+c)^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^2-2*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elli
pticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*cos(d*x+c)^2*a*b+4*B*Ell
ipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*cos(d*x+c)^2*((b+a*cos(
d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*b^2+3*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*(1/(1+cos(d*x+c)))^(1/2)-2*B*
cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b*(1/(1+cos(d*x+c)))^(1/2)+4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b*(1/(1+c
os(d*x+c)))^(1/2)-4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^2*(1/(1+cos(d*x+c)))^(1/2)-3*B*cos(d*x+c)^2*((a-b)/(a
+b))^(1/2)*a^2*(1/(1+cos(d*x+c)))^(1/2)+3*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b*(1/(1+cos(d*x+c)))^(1/2)-2*B*
cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^2*(1/(1+cos(d*x+c)))^(1/2)+4*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2*(1/(1+cos
(d*x+c)))^(1/2)-B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b*(1/(1+cos(d*x+c)))^(1/2)+2*B*((a-b)/(a+b))^(1/2)*(1/(1+co
s(d*x+c)))^(1/2)*b^2)/b^2/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)^3/(1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c
)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+B*sec(d*x+c))/cos(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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