3.380 \(\int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=124 \[ \frac {5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4264, 3818, 3787, 3768, 3771, 2639, 2641} \[ \frac {5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x]

[Out]

(3*EllipticE[(c + d*x)/2, 2])/(a*d) + (5*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Cos[c +
d*x]^(3/2)) - (3*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]]) - Sin[c + d*x]/(d*Cos[c + d*x]^(5/2)*(a + a*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 3768

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

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 3818

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

Rule 4264

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

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx\\ &=-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a}{2}-\frac {5}{2} a \sec (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a}\\ &=\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 a}+\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a}\\ &=\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))}+\frac {5 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {3 \int \sqrt {\cos (c+d x)} \, dx}{2 a}\\ &=\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {5 \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 3.75, size = 338, normalized size = 2.73 \[ \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (10 \cos \left (\frac {1}{2} (c-d x)\right )+8 \cos \left (\frac {1}{2} (3 c+d x)\right )+4 \cos \left (\frac {1}{2} (c+3 d x)\right )+5 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+9 \cos \left (\frac {1}{2} (3 c+5 d x)\right )\right ) \sec \left (\frac {1}{2} (c+d x)\right )}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )-5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+9 \left (1+e^{2 i (c+d x)}\right )\right ) \sec (c+d x)}{\left (-1+e^{2 i c}\right ) d \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])),x]

[Out]

(Cos[(c + d*x)/2]^2*(-1/4*((10*Cos[(c - d*x)/2] + 8*Cos[(3*c + d*x)/2] + 4*Cos[(c + 3*d*x)/2] + 5*Cos[(5*c + 3
*d*x)/2] + 9*Cos[(3*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2])/(d*Cos[c + d*x]^(5/2)) + ((2*I)*Sqrt[2]
*(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))]*Hypergeometr
ic2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x])/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((
2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(3*a*(1 + Sec[c + d*x]))

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)

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maple [B]  time = 6.56, size = 413, normalized size = 3.33 \[ \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 )}\, \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 )}\, \left (10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+44 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-11 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\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/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x)

[Out]

1/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a/sin(1/2*d*x+1/2*c)^3/cos(1/2*d*x+1/2*c)/(4*sin
(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^2-18*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-36*sin(1/2*d*x+1/2*c)^6-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*cos(1/2*d*x+1/2*c)+9*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*cos(1/2*d*x+1/2*c)+44*sin(1/2
*d*x+1/2*c)^4-11*sin(1/2*d*x+1/2*c)^2)/(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 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))),x)

[Out]

int(1/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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