3.6.44 \(\int \frac {(a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [544]

Optimal. Leaf size=208 \[ \frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (A-3 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \]

[Out]

2/3*A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/3*a^2*(A-3*B-5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/3*(
A-C)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d+4*a^2*(A-C)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^2*(2*A+3*B+2*C)*(cos(1/2
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2
)/d

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Rubi [A]
time = 0.30, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4171, 4103, 4082, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {2 a^2 (A-3 B-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*(2*A + 3*B + 2*C)*S
qrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (2*a^2*(A - 3*B - 5*C)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) - (2*(A - C)*Sqrt
[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(3*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

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 3872

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 4082

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

Rule 4103

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

Rule 4171

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

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (4 A+3 B)-\frac {3}{2} a (A-C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {4 \int \frac {(a+a \sec (c+d x)) \left (\frac {3}{4} a^2 (5 A+3 B-C)-\frac {3}{4} a^2 (A-3 B-5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 a^2 (A-3 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {8 \int \frac {\frac {9}{4} a^3 (A-C)+\frac {3}{4} a^3 (2 A+3 B+2 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 a^2 (A-3 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\left (2 a^2 (A-C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (2 a^2 (2 A+3 B+2 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=-\frac {2 a^2 (A-3 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\left (2 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a^2 (2 A+3 B+2 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (A-3 B-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.62, size = 209, normalized size = 1.00 \begin {gather*} \frac {a^2 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (12 i A-12 i C+12 i A \cos (2 (c+d x))-12 i C \cos (2 (c+d x))+8 (2 A+3 B+2 C) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-4 i (A-C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+A \sin (c+d x)+4 C \sin (c+d x)+6 B \sin (2 (c+d x))+12 C \sin (2 (c+d x))+A \sin (3 (c+d x))\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((12*I)*A - (12*I)*C + (12*I)*A*Cos[2*(c + d*x)] - (12*I)*C*Co
s[2*(c + d*x)] + 8*(2*A + 3*B + 2*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - (4*I)*(A - C)*(1 + E^((2*I
)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + A*Sin[c + d*x] + 4*C*Sin[c + d*x]
+ 6*B*Sin[2*(c + d*x)] + 12*C*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(6*d*E^(I*d*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(799\) vs. \(2(238)=476\).
time = 0.13, size = 800, normalized size = 3.85

method	result	size
default	\(\text {Expression too large to display}\)	\(800\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

4/3*a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)
^2+1)/sin(1/2*d*x+1/2*c)^3*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^4+4*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*
sin(1/2*d*x+1/2*c)^2-6*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*B*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-12*C*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*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)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+6*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)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-2*A*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^2-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))+7*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C-2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)
^2-1)^(1/2)*EllipticE(cos(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)/(2*cos
(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.82, size = 229, normalized size = 1.10 \begin {gather*} -\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (A a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(I*sqrt(2)*(2*A + 3*B + 2*C)*a^2*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
I*sqrt(2)*(2*A + 3*B + 2*C)*a^2*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*
sqrt(2)*(A - C)*a^2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) + 3*I*sqrt(2)*(A - C)*a^2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I
*sin(d*x + c))) - (A*a^2*cos(d*x + c)^2 + 3*(B + 2*C)*a^2*cos(d*x + c) + C*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)
))/(d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int A \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \frac {B}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 2 B \sqrt {\sec {\left (c + d x \right )}}\, dx + \int B \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int C \sqrt {\sec {\left (c + d x \right )}}\, dx + \int 2 C \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int C \sec ^{\frac {5}{2}}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A/sec(c + d*x)**(3/2), x) + Integral(2*A/sqrt(sec(c + d*x)), x) + Integral(A*sqrt(sec(c + d*x))
, x) + Integral(B/sqrt(sec(c + d*x)), x) + Integral(2*B*sqrt(sec(c + d*x)), x) + Integral(B*sec(c + d*x)**(3/2
), x) + Integral(C*sqrt(sec(c + d*x)), x) + Integral(2*C*sec(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**(5
/2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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