3.660 \(\int \frac{(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=259 \[ \frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (a^2 C+A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a b d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a d \sqrt{a+b \cos (c+d x)}}+\frac{2 C \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}} \]

[Out]

(-2*(A*b^2 + a^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(a*b*(a^2 - b^2)*d*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]) + (2*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/
(b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a +
b)])/(a*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*
x]])

________________________________________________________________________________________

Rubi [A]  time = 0.702017, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3056, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (a^2 C+A b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a b d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a d \sqrt{a+b \cos (c+d x)}}+\frac{2 C \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(-2*(A*b^2 + a^2*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(a*b*(a^2 - b^2)*d*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]) + (2*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/
(b*d*Sqrt[a + b*Cos[c + d*x]]) + (2*A*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a +
b)])/(a*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*
x]])

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3002

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 \int \frac{\left (\frac{1}{2} A \left (a^2-b^2\right )-\frac{1}{2} a b (A+C) \cos (c+d x)-\frac{1}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 \int \frac{\left (-\frac{1}{2} A b \left (a^2-b^2\right )-\frac{1}{2} a \left (a^2-b^2\right ) C \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a b \left (a^2-b^2\right )}-\frac{\left (A b^2+a^2 C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{a b \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{A \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a}+\frac{C \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{b}-\frac{\left (\left (A b^2+a^2 C\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{a b \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (A b^2+a^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a b \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\left (A \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{a \sqrt{a+b \cos (c+d x)}}+\frac{\left (C \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{b \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (A b^2+a^2 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a b \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}}+\frac{2 A \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{a d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 29.7452, size = 0, normalized size = 0. \[ \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^(3/2), x]

________________________________________________________________________________________

Maple [A]  time = 0.821, size = 539, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2
*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*(-A*b^2-C*a^2)/a/b/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a
^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)
*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)
)*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-2*A/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b
+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),
2,(-2*b/(a-b))^(1/2)))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)

________________________________________________________________________________________

Fricas [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+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a + b \cos{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)/(a + b*cos(c + d*x))**(3/2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c) + a)^(3/2), x)