3.500 \(\int \frac{A+B \sec (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=113 \[ \frac{(A-B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a d}+\frac{(A-3 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A-3 B) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{(A-B) \sin (c+d x)}{d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.239635, antiderivative size = 113, 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, 2978, 2748, 2636, 2639, 2641} \[ \frac{(A-B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(A-3 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A-3 B) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{(A-B) \sin (c+d x)}{d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - 3*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((A - B)*EllipticF[(c + d*x)/2, 2])/(a*d) - ((A - 3*B)*Sin[c + d
*x])/(a*d*Sqrt[Cos[c + d*x]]) + ((A - B)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(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 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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

Mathematica [C]  time = 6.6631, size = 1240, normalized size = 10.97 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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))*Si
n[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)]*Sq
rt[(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*Hy
pergeometric2F1[-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*B - A*Cos[c] + B*Cos[c])*Csc[c/2]*Sec[c/2
]*Sec[c])/d + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/d + (4*B*Sec[c]*Sec[c + d*x
]*Sin[d*x])/d))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])) - (A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricP
FQ[{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]]]*Sqr
t[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*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[C
ot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])
)

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Maple [A]  time = 4.232, size = 318, normalized size = 2.8 \begin{align*} -{\frac{1}{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{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\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}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( A{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -A{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -B{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +3\,B{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) +2\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( A-3\,B \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-\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 ( A-5\,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} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3} \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \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((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*(-cos(1/2*d*x+1/2*c)*(sin(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*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*Elliptic
E(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)*(A-3*B)*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)*(A-5*B)*sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/sin
(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2-1)/(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{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \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+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{2}}, x\right ) \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+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c)^2*sec(d*x + c) + a*cos(d*x + c)^2), 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+B*sec(d*x+c))/cos(d*x+c)**(3/2)/(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{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \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+a*sec(d*x+c)),x, algorithm="giac")

[Out]

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