3.215 \(\int \frac {1}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=195 \[ -\frac {13 \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {13 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {49 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {8 \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

-1/5*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^3-8/15*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^2-
13/6*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a^3+a^3*sec(d*x+c))+49/10*(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^3/d-13/6*(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^3/d

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Rubi [A]  time = 0.32, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3817, 4020, 3787, 3771, 2639, 2641} \[ -\frac {13 \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {13 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {49 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {8 \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(49*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - (13*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x
])^3) - (8*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (13*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(6*d*(a^3 + a^3*Sec[c + d*x]))

Rule 2639

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

Rule 2641

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

Rule 3771

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 3787

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 3817

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

Rule 4020

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-\frac {11 a}{2}+\frac {5}{2} a \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {41 a^2}{2}+12 a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \frac {-\frac {147 a^3}{4}+\frac {65}{4} a^3 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {13 \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}+\frac {49 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (13 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}+\frac {\left (49 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {49 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {13 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 2.07, size = 386, normalized size = 1.98 \[ \frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-\frac {1}{32} \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (1134 \cos \left (\frac {1}{2} (c-d x)\right )+1071 \cos \left (\frac {1}{2} (3 c+d x)\right )+923 \cos \left (\frac {1}{2} (c+3 d x)\right )+694 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+470 \cos \left (\frac {1}{2} (3 c+5 d x)\right )+265 \cos \left (\frac {1}{2} (7 c+5 d x)\right )+117 \cos \left (\frac {1}{2} (5 c+7 d x)\right )+30 \cos \left (\frac {1}{2} (9 c+7 d x)\right )\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}+\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (147 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+65 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+147 \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{15 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^6*(((2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(147*(1 + E^((2*I)*(c +
d*x))) + 147*(-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))] + 65*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))) - ((1134*Cos[(c - d*x)/2] + 1071*Cos[(3*c + d*x)
/2] + 923*Cos[(c + 3*d*x)/2] + 694*Cos[(5*c + 3*d*x)/2] + 470*Cos[(3*c + 5*d*x)/2] + 265*Cos[(7*c + 5*d*x)/2]
+ 117*Cos[(5*c + 7*d*x)/2] + 30*Cos[(9*c + 7*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5*Sqrt[Sec[c + d*x]])
/32)*Sec[c + d*x]^3)/(15*a^3*d*(1 + Sec[c + d*x])^3)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{4} + 3 \, a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(sec(d*x + c))/(a^3*sec(d*x + c)^4 + 3*a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + a^3*sec(d*x +
c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 4.28, size = 270, normalized size = 1.38 \[ \frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (348 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+130 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+294 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-578 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+264 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-37 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3\right )}{60 a^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*cos(1/2*d*x+1/2*c)^8+130*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+294
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))-578*cos(1/2*d*x+1/2*c)^6+264*cos(1/2*d*x+1/2*c)^4-37*cos(1/2*d*x+1/2*c)^2+3)/(-2*sin(1/2*d*x+1/2*
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5/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 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sec(c + d*x)**(7/2) + 3*sec(c + d*x)**(5/2) + 3*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x))), x)/a**3

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