3.497 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx\)
Optimal. Leaf size=124 \[ \frac{(5 A-3 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a d}-\frac{3 (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(5 A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a d} \]
[Out]
(-3*(A - B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((5*A - 3*B)*EllipticF[(c + d*x)/2, 2])/(3*a*d) + ((5*A - 3*B)*
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) - ((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))
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Rubi [A] time = 0.244492, antiderivative size = 124, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used =
{2954, 2977, 2748, 2639, 2635, 2641} \[ \frac{(5 A-3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}-\frac{3 (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(5 A-3 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
[Out]
(-3*(A - B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((5*A - 3*B)*EllipticF[(c + d*x)/2, 2])/(3*a*d) + ((5*A - 3*B)*
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) - ((A - B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + 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 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 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 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
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{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx &=\int \frac{\cos ^{\frac{3}{2}}(c+d x) (B+A \cos (c+d x))}{a+a \cos (c+d x)} \, dx\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{\int \sqrt{\cos (c+d x)} \left (-\frac{3}{2} a (A-B)+\frac{1}{2} a (5 A-3 B) \cos (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{(5 A-3 B) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{2 a}-\frac{(3 (A-B)) \int \sqrt{\cos (c+d x)} \, dx}{2 a}\\ &=-\frac{3 (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(5 A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{(5 A-3 B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac{3 (A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(5 A-3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{(5 A-3 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}
Mathematica [C] time = 6.53272, size = 1239, normalized size = 9.99 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x]),x]
[Out]
(((-3*I)/4)*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/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*I)/4)*B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/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)*C
os[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])) + (Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x])*((-2*(-A + B)*(1 + 2*Cos[c])*Csc[c])/d +
(4*A*Cos[d*x]*Sin[c])/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*A*Co
s[c]*Sin[d*x])/(3*d)))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) - (5*A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hyperg
eometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/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])) + (B*Cos[c/2
+ (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/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*Se
c[c + d*x]))
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Maple [A] time = 2.127, size = 262, normalized size = 2.1 \begin{align*} -{\frac{1}{3\,ad}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 5\,A{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +9\,A{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -3\,B{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -9\,B{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) -8\,A \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 18\,A-6\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -7\,A+3\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x)
[Out]
-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-8*A*sin(1/2*d*x+1
/2*c)^6+(18*A-6*B)*sin(1/2*d*x+1/2*c)^4+(-7*A+3*B)*sin(1/2*d*x+1/2*c)^2)/a/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*sqrt(cos(d*x + c))/(a*sec(d*x + c) + a), 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(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),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 )} \cos \left (d x + c\right )^{\frac{3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)