3.632 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=329 \[ \frac {2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {2 (A b-a B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \left (a^3 B+2 a^2 A b-5 a b^2 B+2 A b^3\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
[Out]
2/3*a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2)+2/3*(2*A*a^2*b+2*A*b^3+B*a^3-
5*B*a*b^2)*sin(d*x+c)/b/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2/3*(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)/a/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2/3*(3*A*a^2+A*b^2-4*B*a*b)*(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)/a/(a^2-b^2)^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
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Rubi [A] time = 1.04, antiderivative size = 329, normalized size of antiderivative =
1.00, number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\)
= 0.286, Rules used = {2955, 4029, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac {2 \left (2 a^2 A b+a^3 B-5 a b^2 B+2 A b^3\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {2 (A b-a B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (3 a^2 A-4 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]
[Out]
(-2*(A*b - a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a*(a^2 - b^2)*d*S
qrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(3*a^2*A + A*b^2 - 4*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (
2*a*(A*b - a*B)*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*(2*a^2*A*
b + 2*A*b^3 + a^3*B - 5*a*b^2*B)*Sin[c + d*x])/(3*b*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[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 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 4029
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*d^2*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])
^(n - 2))/(b*f*(m + 1)*(a^2 - b^2)), x] - Dist[d/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*
Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*(n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) -
d*B*(a^2*(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a
*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 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 4100
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[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{2} a (A b-a B)-\frac {3}{2} b (A b-a B) \sec (c+d x)+\frac {1}{2} \left (2 a A b+a^2 B-3 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a b \left (3 a^2 A+A b^2-4 a b B\right )+\frac {1}{4} a b \left (4 a A b-a^2 B-3 b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 a b \left (a^2-b^2\right )^2}\\ &=\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left ((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}{3 a \left (a^2-b^2\right )}-\frac {\left (\left (3 a^2 A+A b^2-4 a b B\right ) \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}{3 a \left (a^2-b^2\right )^2}\\ &=\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left ((A b-a B) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 a \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (3 a^2 A+A b^2-4 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left ((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}{3 a \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (3 a^2 A+A b^2-4 a b B\right ) \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}{3 a \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=-\frac {2 (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 )}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (3 a^2 A+A b^2-4 a b B\right ) \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)}}{3 a \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 A b+2 A b^3+a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 16.80, size = 487, normalized size = 1.48 \[ \frac {(a \cos (c+d x)+b)^3 \left (\frac {2 (A b \sin (c+d x)-a B \sin (c+d x))}{3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac {2 \left (3 a^2 A \sin (c+d x)-4 a b B \sin (c+d x)+A b^2 \sin (c+d x)\right )}{3 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}\right )}{d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sec ^{\frac {5}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b)^2 \left (-\left (3 a^2 A-4 a b B+A b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i (a+b) \left (3 a^2 A-4 a b B+A b^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+i a (a+b) (3 a A-a B+A b-3 b B) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{3 a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]
[Out]
((b + a*Cos[c + d*x])^3*((2*(A*b*Sin[c + d*x] - a*B*Sin[c + d*x]))/(3*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) + (
2*(3*a^2*A*Sin[c + d*x] + A*b^2*Sin[c + d*x] - 4*a*b*B*Sin[c + d*x]))/(3*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])))
)/(d*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)) + (2*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x
]^(5/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*((-I)*(a + b)*(3*a^2*A + A*b^2 - 4*a*b*B)*EllipticE[I*ArcSinh[
Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)
] + I*a*(a + b)*(3*a*A + A*b - a*B - 3*b*B)*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c +
d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (3*a^2*A + A*b^2 - 4*a*b*B)*(b + a*Cos[c +
d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(3*a*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(5/2))
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fricas [F] time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^2*sec(d*x + c)^3 +
3*a*b^2*cos(d*x + c)^2*sec(d*x + c)^2 + 3*a^2*b*cos(d*x + c)^2*sec(d*x + c) + a^3*cos(d*x + c)^2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)
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maple [B] time = 2.29, size = 1925, normalized size = 5.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x)
[Out]
2/3/d*(-1+cos(d*x+c))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-3*B*((a-b)/(a+b))^(1/2)*cos(d*x+c
)^2*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+3*A*sin(d*x+c)*cos(d*x+c)*EllipticF((-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)*a^3+3*A*((a-b)/(a+b))^(1/2)*cos(d*
x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+4*B*((a-b)
/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-4*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^2*(1/(1+cos(d*x
+c)))^(1/2)+B*sin(d*x+c)*EllipticF((-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)*a^2*b+4*B*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ell
ipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+3*A*sin(d*x+c)*EllipticF((-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
)*a^2*b-A*sin(d*x+c)*EllipticF((-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)*a*b^2-3*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti
cE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-B*((a-b)/(a+b))^(1/2)*a^3*(1/(1+
cos(d*x+c)))^(1/2)-A*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(1/2)+B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*(
1/(1+cos(d*x+c)))^(1/2)+3*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^3*(1/(1+cos(d*x+c)))^(1/2)-3*A*((a-b)/(a+b))^(1
/2)*cos(d*x+c)*a^3*(1/(1+cos(d*x+c)))^(1/2)+A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^3*(1/(1+cos(d*x+c)))^(1/2)-2*A*
((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+A*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)-B*((a-
b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+4*B*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)-A*sin(d*
x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-
(a+b)/(a-b))^(1/2))*b^3-3*B*sin(d*x+c)*EllipticF((-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)*a*b^2-3*B*sin(d*x+c)*cos(d*x+c)*EllipticF((-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)*a^2*b+4*B
*sin(d*x+c)*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(
1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-3*A*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticE((-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^3+B*sin(d*x+c)*cos(d*x+
c)*EllipticF((-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)*a^3-A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-A*sin(d*x+c)*cos(d*x+c
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+
b)/(a-b))^(1/2))*a*b^2-A*sin(d*x+c)*cos(d*x+c)*EllipticF((-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)*a^2*b)*cos(d*x+c)^(1/2)*((a-b)/(a+b))^(1/2)*(1/(
1+cos(d*x+c)))^(1/2)/a/(a+b)/(a-b)^2/(b+a*cos(d*x+c))^2/sin(d*x+c)^3
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + b/cos(c + d*x))^(5/2)),x)
[Out]
int((A + B/cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + b/cos(c + d*x))^(5/2)), 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)**(3/2)/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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