3.637 \(\int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=187 \[ \frac {2 \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\sec (c+d x)}}+\frac {8 b \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
[Out]
2/3*(a^2-b^2)*(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)*sec(d*x+c)^(1/2)/d/(a+b*sec(d*x+c))^(1/2)+2/3*a*sin(d*x+c)*(a+b*sec(d*x+c
))^(1/2)/d/sec(d*x+c)^(1/2)+8/3*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))*(a+b*sec(d*x+c))^(1/2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
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Rubi [A] time = 0.41, antiderivative size = 187, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used =
{3864, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\sec (c+d x)}}+\frac {8 b \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(3/2),x]
[Out]
(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3
*d*Sqrt[a + b*Sec[c + d*x]]) + (8*b*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*d*Sqrt[
(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[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 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 3864
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[(a*Co
t[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e + f*x]^2, x])/Sqrt[a + b*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[2*n]
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]
Rubi steps
\begin {align*} \int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {1}{3} \int \frac {-4 a b-\left (a^2+3 b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} (4 b) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx-\frac {1}{3} \left (-a^2+b^2\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {\left (\left (-a^2+b^2\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {\left (\left (-a^2+b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{3 \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 b \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{3 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 d \sqrt {a+b \sec (c+d x)}}+\frac {8 b E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.75, size = 156, normalized size = 0.83 \[ \frac {(a+b \sec (c+d x))^{3/2} \left (2 \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+2 a \sin (c+d x) (a \cos (c+d x)+b)+8 b (a+b) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(3/2),x]
[Out]
((a + b*Sec[c + d*x])^(3/2)*(8*b*(a + b)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a +
b)] + 2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(b + a*Cos[
c + d*x])*Sin[c + d*x]))/(3*d*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(3/2))
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fricas [F] time = 2.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
integral((b*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(3/2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(3/2), x)
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maple [B] time = 1.89, size = 1219, normalized size = 6.52 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x)
[Out]
-2/3/d*(cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*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^2-4*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*
x+c)))^(1/2)*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+3*cos(d*x+c)*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)*(1/(1+cos(d*x+c)))^(1/2)*b^2+4*cos(d*x+
c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(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+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b-4*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(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))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(
a+b))^(1/2)*b^2+cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2+EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*a^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-4*
EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+c
os(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*sin(d*x+c
)+4*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b*((b+a*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-4*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin
(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d
*x+c)+5*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b-((a-b)/(a+b))^(1/2)*a^2*cos(d*x+c)-4*cos(d*x+c)*((a-b)/(a+b))^(1/
2)*a*b+4*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^2-a*b*((a-b)/(a+b))^(1/2)-4*b^2*((a-b)/(a+b))^(1/2))*((b+a*cos(d*x+c
))/cos(d*x+c))^(1/2)*cos(d*x+c)^2*(1/cos(d*x+c))^(3/2)/sin(d*x+c)/(b+a*cos(d*x+c))/((a-b)/(a+b))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(3/2),x)
[Out]
int((a + b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))**(3/2)/sec(d*x+c)**(3/2),x)
[Out]
Integral((a + b*sec(c + d*x))**(3/2)/sec(c + d*x)**(3/2), x)
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