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

Optimal. Leaf size=160 \[ \frac{4 a^2 (3 A+2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a^2 (3 A+5 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 B \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{3 d}-\frac{4 a^2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.278166, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4018, 3997, 3787, 3771, 2639, 2641} \[ \frac{2 a^2 (3 A+5 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{4 a^2 (3 A+2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sin (c+d x) \sqrt{\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{3 d}-\frac{4 a^2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 3.07969, size = 310, normalized size = 1.94 \[ \frac{a^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 (A+B \sec (c+d x)) \left (\frac{-3 \csc (c) \cos (d x) (A \cos (2 c)-A-4 B)+6 A \cos (c) \sin (d x)+2 B \tan (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)}-\frac{4 i \sqrt{2} e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^3(c+d x) \left (e^{i d x} \left (\left (-1+e^{2 i c}\right ) (3 A+2 B) \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+B e^{i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )\right )+3 B e^{i c} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{12 d (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E
^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^3*(3*B*E^(I*c)*Hypergeometric2F1[-1/4, 1/2, 3/
4, -E^((2*I)*(c + d*x))] + E^(I*d*x)*((3*A + 2*B)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I
)*(c + d*x))] + B*E^(I*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])))/(E^(I*d*x)*(-1 + E
^((2*I)*c))) + (-3*(-A - 4*B + A*Cos[2*c])*Cos[d*x]*Csc[c] + 6*A*Cos[c]*Sin[d*x] + 2*B*Tan[c + d*x])/Sec[c + d
*x]^(5/2)))/(12*d*(B + A*Cos[c + d*x]))

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Maple [B]  time = 2.067, size = 513, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

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]

-4/3*a^2*(6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+2*B)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A+7*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(2
*sin(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)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+3*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)*El
lipticF(cos(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)+2*B*(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(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)+3*B*(-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)^(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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} \end{align*}

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

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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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+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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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} \end{align*}

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)