3.2.99 \(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [199]

Optimal. Leaf size=140 \[ -\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {5 \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \]

[Out]

5/3*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)-sin(d*x+c)/d/(a+a*sec(d*x+c))/sec(d*x+c)^(1/2)-3*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d+5/3*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/
2)/a/d

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Rubi [A]
time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3904, 3872, 3854, 3856, 2720, 2719} \begin {gather*} \frac {5 \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}+\frac {5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

(-3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (5*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(d*Sqr
t[Sec[c + d*x]]*(a + a*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[Cot[e + f*x
]*((d*Csc[e + f*x])^n/(f*(a + b*Csc[e + f*x]))), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}-\frac {\int \frac {-\frac {5 a}{2}+\frac {3}{2} a \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}-\frac {3 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a}+\frac {5 \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac {5 \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}+\frac {5 \int \sqrt {\sec (c+d x)} \, dx}{6 a}-\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a}\\ &=-\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {5 \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {5 \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.03, size = 318, normalized size = 2.27 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (9 \left (1+e^{2 i (c+d x)}\right )+9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+2 \sqrt {\sec (c+d x)} \left (3 (2+\cos (2 c)) \cos (d x) \csc (c)+\cos (2 d x) \sin (2 c)-3 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-6 \cos (c) \sin (d x)+\cos (2 c) \sin (2 d x)-3 \tan \left (\frac {c}{2}\right )\right )\right )}{3 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])),x]

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(9*(1 + E^((
2*I)*(c + d*x))) + 9*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2
*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/
2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + 2*Sqrt[Sec[c + d*x]]*(3*(2 + Cos[2*c])*
Cos[d*x]*Csc[c] + Cos[2*d*x]*Sin[2*c] - 3*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] - 6*Cos[c]*Sin[d*x] + Cos[2*c
]*Sin[2*d*x] - 3*Tan[c/2])))/(3*a*d*(1 + Sec[c + d*x]))

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Maple [A]
time = 0.07, size = 215, normalized size = 1.54

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/3/a*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2)))-8*sin(1/2*d*x+1/2*c)^6+18*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*sec(d*x + c)^(3/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.84, size = 207, normalized size = 1.48 \begin {gather*} -\frac {5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(5*(I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(-I
*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(I*sqrt(2)*co
s(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*
(-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d
*x + c))) - 2*(2*cos(d*x + c)^2 + 5*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c)),x)

[Out]

Integral(1/(sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x)/a

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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(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*sec(d*x + c)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(3/2)),x)

[Out]

int(1/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(3/2)), x)

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