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

Optimal. Leaf size=261 \[ \frac{(33 A-13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}+\frac{(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{7 (17 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^3} \]

[Out]

(-7*(17*A - 7*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((33*A - 13*B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((33*A - 13*B)*Sin[c + d*x])/(6*a
^3*d*Sqrt[Sec[c + d*x]]) - ((A - B)*Sin[c + d*x])/(5*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) - ((2*A - B)
*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) - (7*(17*A - 7*B)*Sin[c + d*x])/(30*d*Sqrt[Se
c[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.553005, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4020, 3787, 3769, 3771, 2641, 2639} \[ \frac{(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(33 A-13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{7 (17 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*(17*A - 7*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((33*A - 13*B)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + ((33*A - 13*B)*Sin[c + d*x])/(6*a
^3*d*Sqrt[Sec[c + d*x]]) - ((A - B)*Sin[c + d*x])/(5*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) - ((2*A - B)
*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) - (7*(17*A - 7*B)*Sin[c + d*x])/(30*d*Sqrt[Se
c[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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 3769

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

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 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{A+B \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (13 A-3 B)-\frac{7}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a^2 (23 A-8 B)-\frac{25}{2} a^2 (2 A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{\frac{15}{4} a^3 (33 A-13 B)-\frac{21}{4} a^3 (17 A-7 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(33 A-13 B) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{4 a^3}-\frac{(7 (17 A-7 B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=\frac{(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(33 A-13 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}-\frac{\left (7 (17 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{7 (17 A-7 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}+\frac{(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left ((33 A-13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac{7 (17 A-7 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}+\frac{(33 A-13 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}+\frac{(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt{\sec (c+d x)}}-\frac{(A-B) \sin (c+d x)}{5 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac{(2 A-B) \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt{\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.89894, size = 377, normalized size = 1.44 \[ \frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) (A+B \sec (c+d x)) \left (7 i (17 A-7 B) e^{-\frac{3}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^5 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+160 (33 A-13 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \cos (c+d x) \left (-210 i (17 A-7 B) \cos \left (\frac{1}{2} (c+d x)\right )-105 i (17 A-7 B) \cos \left (\frac{3}{2} (c+d x)\right )+352 A \sin \left (\frac{1}{2} (c+d x)\right )+545 A \sin \left (\frac{3}{2} (c+d x)\right )+227 A \sin \left (\frac{5}{2} (c+d x)\right )+10 A \sin \left (\frac{7}{2} (c+d x)\right )-357 i A \cos \left (\frac{5}{2} (c+d x)\right )-142 B \sin \left (\frac{1}{2} (c+d x)\right )-205 B \sin \left (\frac{3}{2} (c+d x)\right )-87 B \sin \left (\frac{5}{2} (c+d x)\right )+147 i B \cos \left (\frac{5}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\sec (c+d x)+1)^3 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(Cos[d*x] + I*Sin[d*x])*(160*(33*A - 13*B)*Cos[(c +
d*x)/2]^5*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((7*I)*(17*A - 7*B)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 +
E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(((3*I)/2)*(c + d*x)) + 2*Cos[c
+ d*x]*((-210*I)*(17*A - 7*B)*Cos[(c + d*x)/2] - (105*I)*(17*A - 7*B)*Cos[(3*(c + d*x))/2] - (357*I)*A*Cos[(5
*(c + d*x))/2] + (147*I)*B*Cos[(5*(c + d*x))/2] + 352*A*Sin[(c + d*x)/2] - 142*B*Sin[(c + d*x)/2] + 545*A*Sin[
(3*(c + d*x))/2] - 205*B*Sin[(3*(c + d*x))/2] + 227*A*Sin[(5*(c + d*x))/2] - 87*B*Sin[(5*(c + d*x))/2] + 10*A*
Sin[(7*(c + d*x))/2])))/(120*a^3*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 2.198, size = 465, normalized size = 1.8 \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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(160*A*cos(1/2*d*x+1/2*c)^10+468*A*cos(1/2*d
*x+1/2*c)^8+330*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))+714*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2
+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-348*B*cos(1/2*d*x+1/2*c)^8-130*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-294*B*cos(1/2*d*x
+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-1058*A*cos(1/2*d*x+1/2*c)^6+578*B*cos(1/2*d*x+1/2*c)^6+474*A*cos(1/2*d*x+1/2*c)^4-264*B*cos(1/2*d*x+1/2*c)^4-
47*A*cos(1/2*d*x+1/2*c)^2+37*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/cos(1/2*d*x+1/2*c)^5/(-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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

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{\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{5} + 3 \, a^{3} \sec \left (d x + c\right )^{4} + 3 \, a^{3} \sec \left (d x + c\right )^{3} + a^{3} \sec \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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^3*sec(d*x + c)^5 + 3*a^3*sec(d*x + c)^4 + 3*a^3*sec(d*x +
c)^3 + a^3*sec(d*x + c)^2), x)

________________________________________________________________________________________

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))/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{3} \sec \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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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