3.584 \(\int \frac{A+B \sec (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=197 \[ -\frac{(A b-a B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a d \left (a^2-b^2\right )}-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2}+\frac{a (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
[Out]
-(((A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(b*(a^2 - b^2)*d)) - ((A*b - a*B)*EllipticF[(c + d*x)/2, 2])/(a*(a^2
- b^2)*d) + ((a^2*A*b + A*b^3 + a^3*B - 3*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a - b)*b*(a
+ b)^2*d) + (a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
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Rubi [A] time = 0.666893, antiderivative size = 197, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used =
{2954, 3000, 3059, 2639, 3002, 2641, 2805} \[ -\frac{(A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2}+\frac{a (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]
[Out]
-(((A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(b*(a^2 - b^2)*d)) - ((A*b - a*B)*EllipticF[(c + d*x)/2, 2])/(a*(a^2
- b^2)*d) + ((a^2*A*b + A*b^3 + a^3*B - 3*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a - b)*b*(a
+ b)^2*d) + (a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*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 3000
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^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && RationalQ[m] && 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 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 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 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 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{A+B \sec (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\int \frac{B+A \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2} \, dx\\ &=\frac{a (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac{\int \frac{\frac{1}{2} \left (-a A b-a^2 B+2 b^2 B\right )+b (A b-a B) \cos (c+d x)+\frac{1}{2} a (A b-a B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} a \left (a A b+a^2 B-2 b^2 B\right )-\frac{1}{2} a b (A b-a B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b \left (a^2-b^2\right )}-\frac{(A b-a B) \int \sqrt{\cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac{(A b-a B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}+\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}-\frac{(A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a (a-b) b (a+b)^2 d}+\frac{a (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.7206, size = 276, normalized size = 1.4 \[ \frac{\frac{-\frac{2 (A b-a B) \sin (c+d x) \left (2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}+\frac{4 b (a B-A b) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{2 \left (3 a^2 B+a A b-4 b^2 B\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b) (a+b)}-\frac{4 a (a B-A b) \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 b d} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^2),x]
[Out]
((-4*a*(-(A*b) + a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) + ((2*(a*A*b + 3*a^2
*B - 4*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (4*b*(-(A*b) + a*B)*(2*EllipticF[(c + d*x)/
2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a - (2*(A*b - a*B)*(-2*a*b*EllipticE[ArcSin[
Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] - (a^2 - 2*b^2)*EllipticPi[-(
a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(4*b*d)
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Maple [B] time = 4.72, size = 715, normalized size = 3.6 \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)^(3/2)/(a+b*sec(d*x+c))^2,x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(-A*b+B*a)/a*(a^2/b/(a^2-b^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)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(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))-1/2*a/b/(a^2-b^2)
*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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)*EllipticPi(cos(1/2*d*x+1/2*c)
,2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-2*A/(a
^2-a*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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)^(3/2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
[Out]
Timed out
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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+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
[Out]
Timed out
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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)**(3/2)/(a+b*sec(d*x+c))**2,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{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)^(3/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*cos(d*x + c)^(3/2)), x)