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

Optimal. Leaf size=153 \[ \frac{(A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a d}+\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}+\frac{(A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.186229, antiderivative size = 153, 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 = {4019, 3787, 3771, 2641, 3768, 2639} \[ \frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}-\frac{(A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}+\frac{(A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

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 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 3768

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

Rubi steps

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

Mathematica [C]  time = 4.37199, size = 420, normalized size = 2.75 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (-2 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+12 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+6 \sqrt{2} B \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )-12 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-6 \sqrt{\sec (c+d x)} \left (2 (B-A) \tan \left (\frac{1}{2} (c+d x)\right )+2 (A-3 B) \csc (c) \cos (d x)\right )\right )}{6 a d (\sec (c+d x)+1) (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((-2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1
+ E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (6*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c
+ d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I
)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 12*A*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 12*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 6*S
qrt[Sec[c + d*x]]*(2*(A - 3*B)*Cos[d*x]*Csc[c] + 2*(-A + B)*Tan[(c + d*x)/2])))/(6*a*d*(B + A*Cos[c + d*x])*(1
+ Sec[c + d*x]))

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Maple [A]  time = 3.819, size = 318, normalized size = 2.1 \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{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}}\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{\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(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)*(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)*(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{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \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(sec(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)*sec(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 \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \sqrt{\sec \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(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*sqrt(sec(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(sec(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 )} \sec \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(sec(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)*sec(d*x + c)^(3/2)/(a*sec(d*x + c) + a), x)