3.744 \(\int \frac{\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=279 \[ \frac{2 (A b-C (2 a+b)) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{b^2 d \sqrt{a+b}}-\frac{2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (2 a^2 C+A b^2-b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b^3 d \sqrt{a+b}} \]

[Out]

(-2*(A*b^2 + 2*a^2*C - b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^3*Sqrt[a + b]*d) + (2*(
A*b - (2*a + b)*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[
(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*Sqrt[a + b]*d) - (2*(A*b^2 + a^2
*C)*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

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Rubi [A]  time = 0.3914, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4081, 4005, 3832, 4004} \[ -\frac{2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (2 a^2 C+A b^2-b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b^3 d \sqrt{a+b}}+\frac{2 (A b-C (2 a+b)) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{b^2 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b^2 + 2*a^2*C - b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^3*Sqrt[a + b]*d) + (2*(
A*b - (2*a + b)*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[
(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*Sqrt[a + b]*d) - (2*(A*b^2 + a^2
*C)*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]])

Rule 4081

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

Rule 4005

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

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

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

Mathematica [A]  time = 18.757, size = 541, normalized size = 1.94 \[ \frac{4 \sqrt{2} \sqrt{\frac{\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left ((a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2} \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (b (C (b-2 a)-A b) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (2 a^2 C+A b^2-b^2 C\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )+\left (2 a^2 C+A b^2-b^2 C\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)\right )}{b^2 d \left (b^2-a^2\right ) \sqrt{\frac{1}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac{(a \cos (c+d x)+b)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\frac{4 \left (a^2 C \sin (c+d x)+A b^2 \sin (c+d x)\right )}{b \left (b^2-a^2\right ) (a \cos (c+d x)+b)}-\frac{4 \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x)}{b^2 \left (b^2-a^2\right )}\right )}{d (a+b \sec (c+d x))^{3/2} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((-4*(A*b^2 + 2*a^2*C - b^2*C)*Sin[c + d*x])/(b^2*(-a^2 + b^2))
+ (4*(A*b^2*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(b*(-a^2 + b^2)*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2
*c + 2*d*x])*(a + b*Sec[c + d*x])^(3/2)) + (4*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*(b + a*Cos[c + d
*x])*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((a
+ b)*((A*b^2 + 2*a^2*C - b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-(A*b) + (-2*a + b)*
C)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*
Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x] + (A*b^2 + 2*a^2*C - b^2*C)*Cos[c + d*x]*(b + a*Cos[c
+ d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(b^2*(-a^2 + b^2)*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*(A + 2*C + A*C
os[2*c + 2*d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.504, size = 2272, normalized size = 8.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/b^2/(a+b)/(a-b)*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3*sin(d*x+
c)+A*b^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*Elliptic
E((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))-C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-C*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)+C*b^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c
))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))-C*b^3*(cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))-C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x
+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+2*C*a^3*(cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))-C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1
/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)*sin(d*x+c)*a*b^2+C*cos(d*x+c)^2*a*b^2
-2*C*cos(d*x+c)^2*a^3-A*cos(d*x+c)*b^3+a^2*b*C-A*cos(d*x+c)^2*a*b^2+A*cos(d*x+c)*a*b^2-2*C*cos(d*x+c)*a^2*b+A*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3+2*C*a^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(
1/2))+A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*
x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)-A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d
*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)+2*C*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(
d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)-2*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(
cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*sin(d*x+c)+2*C*cos(d*x+c)
*a^3+A*b^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*cos(d*
x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a-A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*
(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*co
s(d*x+c)*a*b^2+2*C*a^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*sin(d
*x+c)*cos(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b-2*C*(cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*sin(d*x+c)*cos(d*x+c)*a^2*b+A*cos(d*x+c)^2*b^3+C*cos(d*x+c)^2*a^2*b-C*cos(d*x+c)*a*b^2-A*sin(d*x+c)*cos(d*x+c
)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/
sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-C*b^3-C*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)+C*cos(d*x+c)*b^3)/(b+a*cos(d*x+c))/sin(d*x+c)

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Maxima [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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^3 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c)
+ a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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