3.8.51 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [751]

Optimal. Leaf size=314 \[ -\frac {4 (a-b) b \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b} (a+2 b) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]
time = 0.32, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4307, 2881, 3077, 2895, 3073} \begin {gather*} -\frac {4 b (a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (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 (\text {ArcSin}\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^3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b} (a+2 b) \sqrt {\cos (c+d x)} \csc (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 (\text {ArcSin}\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^2 d \sqrt {\sec (c+d x)}}+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(a - b)*b*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[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^3*d*Sqrt[Sec[c + d*x]]) + (2*Sqrt[a + b]*(a + 2*b)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[Ar
cSin[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^2*d*Sqrt[Sec[c + d*x]]) + (2*Sqrt[a + b*Cos[c + d*x]]*S
ec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d)

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2895

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

Rule 3073

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]/(f*b*c^2))*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)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ
[A, B] && PosQ[(c + d)/b]

Rule 3077

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 4307

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-b+\frac {1}{2} a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a}\\ &=\frac {2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {\left (2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a}+\frac {\left ((a+2 b) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {4 (a-b) b \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (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^3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b} (a+2 b) \sqrt {\cos (c+d x)} \csc (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^2 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}\\ \end {align*}

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Mathematica [A]
time = 13.11, size = 322, normalized size = 1.03 \begin {gather*} \frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+a (a-2 b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+b \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {4 b \sin (c+d x)}{3 a^2}+\frac {2 \tan (c+d x)}{3 a}\right )}{d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*b*(a + b)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c
+ d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + a*(a - 2*b)*Sqrt
[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan
[(c + d*x)/2]], (-a + b)/(a + b)] + b*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/
(3*a^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-
4*b*Sin[c + d*x])/(3*a^2) + (2*Tan[c + d*x])/(3*a)))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(890\) vs. \(2(280)=560\).
time = 0.32, size = 891, normalized size = 2.84

method	result	size
default	\(\text {Expression too large to display}\)	\(891\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/d*(cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2-2*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b
))^(1/2))*a*b+2*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+
b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b+2*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*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)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2-2*cos(d*x+c)*sin(d*x+c)*(cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-
(a-b)/(a+b))^(1/2))*a*b+2*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b+2*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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+cos(d*x+c)^3*a*b-2*cos(d*x+c)^3*b^2+cos(d*x+c)^2*a^2-2*cos(d*x+c)^2*a*b+2*cos(d*x+
c)^2*b^2+cos(d*x+c)*a*b-a^2)*cos(d*x+c)*(1/cos(d*x+c))^(5/2)/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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