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

Optimal. Leaf size=256 \[ \frac{B \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{(2 A b-a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{b d \sqrt{\cos (c+d x)}}-\frac{B \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.888688, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 13, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.371, Rules used = {2955, 4033, 4109, 3859, 2807, 2805, 3862, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{(2 A b-a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{B \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{b d \sqrt{\cos (c+d x)}}+\frac{B \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{B \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2955

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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 4033

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

Rule 4109

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)
]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[
A, Int[1/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2
 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

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]

Rule 3862

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[1/a,
 Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[b/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b
*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
 b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
 d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
 b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

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

Mathematica [C]  time = 32.2606, size = 51168, normalized size = 199.88 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.376, size = 776, normalized size = 3. \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))^(1/2),x)

[Out]

1/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(2*A*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(
1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b-4*A*cos(d*x+c)*sin(d*x+c
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a
+b)/(a-b),I/((a-b)/(a+b))^(1/2))*b-B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*(1/(cos(d*x+c)+1))^(1/2)-2*B*cos(d*x+c
)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin
(d*x+c),(-(a+b)/(a-b))^(1/2))*a+2*B*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elli
pticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a+B*cos(d*x+c)*sin(d*
x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*a-B*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+
cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b+B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*(1/(cos(
d*x+c)+1))^(1/2)-B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b*(1/(cos(d*x+c)+1))^(1/2)+B*((a-b)/(a+b))^(1/2)*b*(1/(cos(d
*x+c)+1))^(1/2))/b/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)+1))^(1/2)/cos(d*x+c)^(1/2)/sin(d*x+c)

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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}{\sqrt{b \sec \left (d x + c\right ) + a} \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))^(1/2),x, algorithm="maxima")

[Out]

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

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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))^(1/2),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))**(1/2),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}{\sqrt{b \sec \left (d x + c\right ) + a} \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))^(1/2),x, algorithm="giac")

[Out]

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