3.187 \(\int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=199 \[ \frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {4 a^2 (5 A+4 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (5 A+4 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

[Out]

2/15*a^2*(5*A+7*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*B*sec(d*x+c)^(3/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d+4/5*
a^2*(5*A+4*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^2*(5*A+4*B)*(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)/d+4/3*a^2*(2*A+B)*(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)/d

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Rubi [A]  time = 0.30, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4018, 3997, 3787, 3771, 2641, 3768, 2639} \[ \frac {2 a^2 (5 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {4 a^2 (5 A+4 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {4 a^2 (5 A+4 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(-4*a^2*(5*A + 4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(2*A + B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (4*a^2*(5*A + 4*B)*Sqrt[Sec[c + d*x]]
*Sin[c + d*x])/(5*d) + (2*a^2*(5*A + 7*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*B*Sec[c + d*x]^(3/2)*(a
^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(5*d)

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 3997

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

Rule 4018

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

Rubi steps

\begin {align*} \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x)) \left (\frac {1}{2} a (5 A+B)+\frac {1}{2} a (5 A+7 B) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {4}{15} \int \sqrt {\sec (c+d x)} \left (\frac {5}{2} a^2 (2 A+B)+\frac {3}{2} a^2 (5 A+4 B) \sec (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{3} \left (2 a^2 (2 A+B)\right ) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (2 a^2 (5 A+4 B)\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {4 a^2 (5 A+4 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {1}{5} \left (2 a^2 (5 A+4 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 (2 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (2 A+B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (5 A+4 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {1}{5} \left (2 a^2 (5 A+4 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {4 a^2 (5 A+4 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (2 A+B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (5 A+4 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (5 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 6.57, size = 321, normalized size = 1.61 \[ \frac {a^2 e^{-i c} \left (-1+e^{2 i c}\right ) \csc (c) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 (A+B \sec (c+d x)) \left (2 (5 A+4 B) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-10 i (2 A+B) \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-30 A e^{i (c+d x)}-60 A e^{3 i (c+d x)}-5 A e^{4 i (c+d x)}-30 A e^{5 i (c+d x)}+5 A-18 B e^{i (c+d x)}-54 B e^{3 i (c+d x)}-10 B e^{4 i (c+d x)}-24 B e^{5 i (c+d x)}+10 B\right )}{60 d \left (1+e^{2 i (c+d x)}\right )^2 \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

(a^2*(-1 + E^((2*I)*c))*Csc[c]*(5*A + 10*B - 30*A*E^(I*(c + d*x)) - 18*B*E^(I*(c + d*x)) - 60*A*E^((3*I)*(c +
d*x)) - 54*B*E^((3*I)*(c + d*x)) - 5*A*E^((4*I)*(c + d*x)) - 10*B*E^((4*I)*(c + d*x)) - 30*A*E^((5*I)*(c + d*x
)) - 24*B*E^((5*I)*(c + d*x)) - (10*I)*(2*A + B)*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[Cos[c + d*x]]*EllipticF[(c +
d*x)/2, 2] + 2*(5*A + 4*B)*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -
E^((2*I)*(c + d*x))])*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/(60*d*E^(I*c)*(1 + E^((2*I
)*(c + d*x)))^2*(B + A*Cos[c + d*x])*Sec[c + d*x]^(5/2))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B a^{2} \sec \left (d x + c\right )^{3} + {\left (A + 2 \, B\right )} a^{2} \sec \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} a^{2} \sec \left (d x + c\right ) + A a^{2}\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)

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maple [B]  time = 13.12, size = 743, normalized size = 3.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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)/(-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)
)+8*(1/4*A+1/2*B)*(-1/6*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+1/3*(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+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+8*(1/2*A+1/4*B)*(-(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))+2*(-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)*sin(1/2*d*x+1/2*c)^2
)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/5*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(
1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)
^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8*sin(1/2*d*x+1/2*c)^2*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))/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 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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