3.3.24 \(\int \frac {(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [224]

Optimal. Leaf size=253 \[ -\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (35 A+13 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

8/105*a^3*(70*A+53*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/7*C*(a+a*sec(d*x+c))^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d+12/
35*C*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d+2/15*(5*A+7*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)*sec
(d*x+c)^(1/2)/d-4/5*a^3*(5*A+7*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+4/21*a^3*(35*A+13*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 = 0.37, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4174, 4103, 4082, 3872, 3856, 2719, 2720} \begin {gather*} \frac {8 a^3 (70 A+53 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d}+\frac {2 (5 A+7 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {12 C \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 \sec (c+d x)+a^2\right )^2}{35 a d}+\frac {2 C \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*(5*A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(35*A + 13
*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (8*a^3*(70*A + 53*C)*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(105*d) + (2*C*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d) + (12*C*Sqr
t[Sec[c + d*x]]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*a*d) + (2*(5*A + 7*C)*Sqrt[Sec[c + d*x]]*(a^3 + a
^3*Sec[c + d*x])*Sin[c + d*x])/(15*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 4174

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

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx &=\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {2 \int \frac {(a+a \sec (c+d x))^3 \left (\frac {1}{2} a (7 A-C)+3 a C \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{7 a}\\ &=\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {4 \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{4} a^2 (35 A-11 C)+\frac {7}{4} a^2 (5 A+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{35 a}\\ &=\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8 \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{4} a^3 (35 A-41 C)+\frac {1}{2} a^3 (70 A+53 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{105 a}\\ &=\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {16 \int \frac {-\frac {21}{8} a^4 (5 A+7 C)+\frac {5}{8} a^4 (35 A+13 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{105 a}\\ &=\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {1}{5} \left (2 a^3 (5 A+7 C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} \left (2 a^3 (35 A+13 C)\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {1}{5} \left (2 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (2 a^3 (35 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {4 a^3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (35 A+13 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {8 a^3 (70 A+53 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 C \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {12 C \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 a d}+\frac {2 (5 A+7 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 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 = 3.74, size = 280, normalized size = 1.11 \begin {gather*} \frac {a^3 e^{-i d x} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-630 i A-882 i C-840 i A \cos (2 (c+d x))-1176 i C \cos (2 (c+d x))-210 i A \cos (4 (c+d x))-294 i C \cos (4 (c+d x))+80 (35 A+13 C) \cos ^{\frac {7}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+14 i (5 A+7 C) e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{7/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+70 A \sin (c+d x)+380 C \sin (c+d x)+630 A \sin (2 (c+d x))+840 C \sin (2 (c+d x))+70 A \sin (3 (c+d x))+260 C \sin (3 (c+d x))+315 A \sin (4 (c+d x))+294 C \sin (4 (c+d x))\right )}{420 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])*((-630*I)*A - (882*I)*C - (840*I)*A*Cos[2*(c + d*x)] - (1176*I
)*C*Cos[2*(c + d*x)] - (210*I)*A*Cos[4*(c + d*x)] - (294*I)*C*Cos[4*(c + d*x)] + 80*(35*A + 13*C)*Cos[c + d*x]
^(7/2)*EllipticF[(c + d*x)/2, 2] + ((14*I)*(5*A + 7*C)*(1 + E^((2*I)*(c + d*x)))^(7/2)*Hypergeometric2F1[1/2,
3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + 70*A*Sin[c + d*x] + 380*C*Sin[c + d*x] + 630*A*Sin[2*(c
+ d*x)] + 840*C*Sin[2*(c + d*x)] + 70*A*Sin[3*(c + d*x)] + 260*C*Sin[3*(c + d*x)] + 315*A*Sin[4*(c + d*x)] +
294*C*Sin[4*(c + d*x)]))/(420*d*E^(I*d*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(986\) vs. \(2(277)=554\).
time = 11.06, size = 987, normalized size = 3.90

method	result	size
default	\(\text {Expression too large to display}\)	\(987\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+4*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6/5*C/sin(1/2*d*x+1/
2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(
1/2*d*x+1/2*c)-12*(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))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*c
os(1/2*d*x+1/2*c)-3*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
)+16*(1/8*A+3/8*C)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+
1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+16*(3/8*A+1/8*C)/sin(1/2*d*x+1/2*c)^2/(2*sin(
1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1
/2*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))))/si
n(1/2*d*x+1/2*c)/(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))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.85, size = 259, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (5 i \, \sqrt {2} {\left (35 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (35 \, A + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\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 ) - 21 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} {\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 (21 \, {\left (15 \, A + 14 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 26 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 63 \, C a^{3} \cos \left (d x + c\right ) + 15 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(5*I*sqrt(2)*(35*A + 13*C)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))
- 5*I*sqrt(2)*(35*A + 13*C)*a^3*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21
*I*sqrt(2)*(5*A + 7*C)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) - 21*I*sqrt(2)*(5*A + 7*C)*a^3*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (21*(15*A + 14*C)*a^3*cos(d*x + c)^3 + 5*(7*A + 26*C)*a^3*cos(d*x + c)^2 + 6
3*C*a^3*cos(d*x + c) + 15*C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7316 deep

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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))^3*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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