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

Optimal. Leaf size=135 \[ \frac {10 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {10 a \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {6 a \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-6/5*a*(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))/d+10/21*a*(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))/d+2/7*a*sin(d*x+c)/d/cos(d*x+c)
^(7/2)+2/5*a*sin(d*x+c)/d/cos(d*x+c)^(5/2)+10/21*a*sin(d*x+c)/d/cos(d*x+c)^(3/2)+6/5*a*sin(d*x+c)/d/cos(d*x+c)
^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4225, 2748, 2636, 2641, 2639} \[ \frac {10 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {10 a \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {6 a \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*EllipticE[(c + d*x)/2, 2])/(5*d) + (10*a*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*Sin[c + d*x])/(7*d*Cos
[c + d*x]^(7/2)) + (2*a*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (10*a*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2))
+ (6*a*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 2636

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

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

\begin {align*} \int \frac {a+a \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\int \frac {a+a \cos (c+d x)}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=a \int \frac {1}{\cos ^{\frac {9}{2}}(c+d x)} \, dx+a \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} (3 a) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} (5 a) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {1}{21} (5 a) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} (3 a) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {10 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {10 a \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 a \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 4.61, size = 294, normalized size = 2.18 \[ \frac {a (\cos (c+d x)+1) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {126 \sec (c) \cos ^3(c+d x) \left (\csc (c) \sqrt {\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \left (3 \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+\cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-2 \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )\right )}{\sqrt {\sec ^2(c)} \sqrt {\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-200 \sin (c) \sqrt {\csc ^2(c)} \cos ^4(c+d x) \sqrt {\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\csc (c) (-85 \cos (2 c+d x)+231 \cos (c+2 d x)+21 \cos (3 c+2 d x)+25 \cos (2 c+3 d x)-25 \cos (4 c+3 d x)+63 \cos (3 c+4 d x)+189 \cos (c)+85 \cos (d x))\right )}{840 d \cos ^{\frac {7}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*((189*Cos[c] + 85*Cos[d*x] - 85*Cos[2*c + d*x] + 231*Cos[c + 2*d*x] +
21*Cos[3*c + 2*d*x] + 25*Cos[2*c + 3*d*x] - 25*Cos[4*c + 3*d*x] + 63*Cos[3*c + 4*d*x])*Csc[c] - 200*Cos[c + d
*x]^4*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[C
ot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] - (126*Cos[c + d*x]^3*Sec[c]*(-2*HypergeometricPFQ[{-1/2, -1/4}, {
3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x
+ ArcTan[Tan[c]]])*Csc[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2
])))/(840*d*Cos[c + d*x]^(7/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a \sec \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 6.01, size = 437, normalized size = 3.24 \[ -\frac {4 \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 )}\, a \left (-\frac {\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 )}}{112 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {5 \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 )}}{84 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {44 \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 )}{105 \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 )}}-\frac {\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 )}}{40 \left (-\frac {1}{2}+\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \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 )}}-\frac {3 \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 (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{10 \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 )}}\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-1/112*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^4-5/84*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+44/105*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
-1/40*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^3-3/
5*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-3/10*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(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 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.38, size = 87, normalized size = 0.64 \[ \frac {2\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2)
) + (2*a*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/(7*d*cos(c + d*x)^(7/2)*(sin(c + d*x)^2)^(
1/2))

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

[Out]

Timed out

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