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

Optimal. Leaf size=215 \[ -\frac {2 (A b-a B) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 A \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2*(A*b-B*a)*sin(d*x+c)/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2*A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/a/d/cos
(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2*(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin
(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a/(a^2-b^2)/d/((b+a*cos(d*x+c
))/(a+b))^(1/2)

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Rubi [A]  time = 0.67, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {2955, 4027, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac {2 (A b-a B) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 A \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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 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 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 4027

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 + 1)*(d*Csc[e + f*x])^(n
- 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e +
f*x])^(n - 1)*Simp[d*(n - 1)*(A*b - a*B) + d*(a*A - b*B)*(m + 1)*Csc[e + f*x] - d*(A*b - a*B)*(m + n + 1)*Csc
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m,
-1] && LtQ[0, n, 1]

Rule 4035

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

Rubi steps

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

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Mathematica [C]  time = 10.92, size = 328, normalized size = 1.53 \[ \frac {2 (a \cos (c+d x)+b) \left (\frac {(a B-A b) \sin (c+d x)}{a^2-b^2}+\frac {\left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left ((A b-a B) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i a (a+b) (A-B) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-i (a+b) (a B-A b) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{\left (a^3-a b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}\right )}{d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{b^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.44, size = 564, normalized size = 2.62 \[ \frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (A \sin \left (d x +c \right ) \EllipticF \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, a +A \sin \left (d x +c \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) b -A \sqrt {\frac {a -b}{a +b}}\, \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, b +B \sin \left (d x +c \right ) \EllipticF \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, a -B \EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \sin \left (d x +c \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, a +B \cos \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}\, a \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+A \sqrt {\frac {a -b}{a +b}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, b -B \sqrt {\frac {a -b}{a +b}}\, a \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a -b}{a +b}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}}{d a \left (b +a \cos \left (d x +c \right )\right ) \left (a -b \right ) \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x)

[Out]

2/d*(-1+cos(d*x+c))*(1+cos(d*x+c))^2*(A*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-
(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a+A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+
c))/(a+b))^(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-A*((a-b)/(a+
b))^(1/2)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*b+B*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin
(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a-B*EllipticE((-1+cos(d*x+c))*((a-
b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a+B*
cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*(1/(1+cos(d*x+c)))^(1/2)+A*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*b-B*(
(a-b)/(a+b))^(1/2)*a*(1/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*((a-b)/(a+
b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)/a/(b+a*cos(d*x+c))/(a-b)/sin(d*x+c)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + b/cos(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(3/2)/cos(d*x+c)**(1/2),x)

[Out]

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

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