3.1076 \(\int \cos ^{\frac {7}{2}}(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=80 \[ \frac {2 (5 A+7 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 (5 A+7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d} \]

[Out]

2/21*(5*A+7*C)*(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*c
os(d*x+c)^(5/2)*sin(d*x+c)/d+2/21*(5*A+7*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4066, 3014, 2635, 2641} \[ \frac {2 (5 A+7 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 (5 A+7 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*A + 7*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2
*A*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d)

Rule 2635

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

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 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 4066

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[b^2, Int
[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e + f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 63, normalized size = 0.79 \[ \frac {2 (5 A+7 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt {\cos (c+d x)} (3 A \cos (2 (c+d x))+13 A+14 C)}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*A + 7*C)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(13*A + 14*C + 3*A*Cos[2*(c + d*x)])*Sin[c + d*x
])/(21*d)

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^3*sec(d*x + c)^2 + A*cos(d*x + c)^3)*sqrt(cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.72, size = 285, normalized size = 3.56 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (56 A +28 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-16 A -14 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+7 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{21 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-72
*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(56*A+28*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-16*A-14*C)*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+7*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2)))/(-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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.58, size = 80, normalized size = 1.00 \[ \frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,A\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*cos(c + d*x)^(1/2)*sin(c + d*x))/(3*d) - (2*A*cos(c + d*x)^(9/2
)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(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(cos(d*x+c)**(7/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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