3.614 \(\int \frac{(a+b \cos (c+d x))^{3/2}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=277 \[ \frac{2 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (a-3 b) (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a d}+\frac{8 b (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a d} \]
[Out]
(8*(a - b)*b*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]
])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(3*a*d) +
(2*(a - 3*b)*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[
c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(3
*a*d) + (2*a*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))
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Rubi [A] time = 0.435061, antiderivative size = 277, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{2799, 2998, 2816, 2994} \[ \frac{2 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (a-3 b) (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a d}+\frac{8 b (a-b) \sqrt{a+b} \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(5/2),x]
[Out]
(8*(a - b)*b*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]
])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(3*a*d) +
(2*(a - 3*b)*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[
c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(3
*a*d) + (2*a*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))
Rule 2799
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 - b^2
)), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[c*(a
*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) + (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d*(
b*c - a*d)*(m + n + 1)*Sin[e + f*x]^2, x], x], 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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]
Rule 2998
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]
Rule 2816
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Rule 2994
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^{3/2}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{2 a b+\frac{1}{2} \left (a^2+3 b^2\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{1}{3} ((a-3 b) (a-b)) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx+\frac{1}{3} (4 a b) \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{8 (a-b) b \sqrt{a+b} \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{3 a d}+\frac{2 (a-3 b) (a-b) \sqrt{a+b} \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right ) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (1+\sec (c+d x))}{a-b}}}{3 a d}+\frac{2 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 4.51187, size = 256, normalized size = 0.92 \[ \frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right )} \left (\left (a^2+4 a b+3 b^2\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )-4 b \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} (a+b \cos (c+d x))-4 b (a+b) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )+\frac{2 \sin (c+d x) (a+b \cos (c+d x)) (a+4 b \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)}}{3 d \sqrt{a+b \cos (c+d x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(5/2),x]
[Out]
((2*(a + b*Cos[c + d*x])*(a + 4*b*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*x]^(3/2) + 2*Sqrt[Cos[(c + d*x)/2]^2]*
(-4*b*(a + b)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2
]^2)/(a + b)] + (a^2 + 4*a*b + 3*b^2)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[((a + b*Cos[c
+ d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - 4*b*(a + b*Cos[c + d*x])*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Tan[(c
+ d*x)/2]))/(3*d*Sqrt[a + b*Cos[c + d*x]])
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Maple [B] time = 0.553, size = 1075, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x)
[Out]
-2/3/d*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x
+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2+4*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/
(a+b))^(1/2))*a*b+3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*b^2-4*cos(d*x+c)^2*sin(d*x+c)*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x
+c),(-(a-b)/(a+b))^(1/2))*a*b-4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1
/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2+(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))
/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2+4*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))
/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b+3*cos(d*x+c)*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/si
n(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*b^2-4*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos
(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-4*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticE
((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2+a*b*cos(d*x+c)^3+4*cos(d*x+c)^3*b^2+cos(d*x+c)^2*a^2+4*c
os(d*x+c)^2*a*b-4*b^2*cos(d*x+c)^2-5*cos(d*x+c)*a*b-a^2)/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^(3/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((b*cos(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\cos \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((b*cos(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))**(3/2)/cos(d*x+c)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((b*cos(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)