3.510 \(\int \frac{\sqrt{\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=188 \[ -\frac{(13 A-3 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}+\frac{(49 A-9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(13 A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(8 A-3 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

((49*A - 9*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - 3*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - ((A
- B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((8*A - 3*B)*Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[c +
d*x]))

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Rubi [A]  time = 0.547795, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2954, 2977, 2748, 2641, 2639} \[ -\frac{(13 A-3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(49 A-9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(13 A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{6 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac{(8 A-3 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((49*A - 9*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - 3*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - ((A
- B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((8*A - 3*B)*Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((13*A - 3*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(6*d*(a^3 + a^3*Cos[c +
d*x]))

Rule 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 2977

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=\int \frac{\cos ^{\frac{5}{2}}(c+d x) (B+A \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\\ &=-\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (-\frac{5}{2} a (A-B)+\frac{1}{2} a (11 A-B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(8 A-3 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (-\frac{3}{2} a^2 (8 A-3 B)+\frac{1}{2} a^2 (41 A-6 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(8 A-3 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{-\frac{5}{4} a^3 (13 A-3 B)+\frac{3}{4} a^3 (49 A-9 B) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(8 A-3 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(49 A-9 B) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}-\frac{(13 A-3 B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac{(49 A-9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(13 A-3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(A-B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(8 A-3 B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.82387, size = 1415, normalized size = 7.53 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((49*I)/10)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((2*E^((2*I)*d*x)*Hy
pergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2
*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I
)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^(
(2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(
I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-
1 + E^((2*I)*d*x))*Sin[c])))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (((9*I)/10)*B*Cos[c/2 + (d*x)/2]^
6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E
^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E
^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3
*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]
*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos
[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((B +
A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (26*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {
5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt
[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x
- ArcTan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) - (2*B*Cos[c/2 + (d*x
)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A +
B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c
]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*
(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x])*((-4*(29*A - 9*B + 20*A*Cos[c])*Csc[c])/(
5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d
*x)/2]*(-29*A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-14*A*Sin[(d*x)/2] +
9*B*Sin[(d*x)/2]))/(15*d) - (4*(-14*A + 9*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(-A + B)*Sec[c/2 + (d
*x)/2]^4*Tan[c/2])/(5*d)))/(Cos[c + d*x]^(3/2)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

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Maple [B]  time = 2.265, size = 451, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*A*cos(1/2*d*x+1/2*c)^8+130*A*(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
*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))-108*B*cos(1/2*d*x+1/2*c)^8-30*B*(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)^5-54*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-578*A*cos(1/2*d*x+1/2*c)^6+198*B*c
os(1/2*d*x+1/2*c)^6+264*A*cos(1/2*d*x+1/2*c)^4-114*B*cos(1/2*d*x+1/2*c)^4-37*A*cos(1/2*d*x+1/2*c)^2+27*B*cos(1
/2*d*x+1/2*c)^2+3*A-3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^3, x)