3.511 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\)
Optimal. Leaf size=208 \[ \frac {\sqrt {2} (A-B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (35 A-7 B+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 a d}-\frac {4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 B-C) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}} \]
[Out]
(A-B+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)-4/105*(35*A-49*B+37*C)
*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/35*(7*B-C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*C*sec(d
*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*(35*A-7*B+31*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d
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Rubi [A] time = 0.64, antiderivative size = 208, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used =
{4088, 4021, 4010, 4001, 3795, 203} \[ \frac {\sqrt {2} (A-B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (35 A-7 B+31 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 a d}-\frac {4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 (7 B-C) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*(35*A
- 49*B + 37*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(7*B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(35
*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A - 7
*B + 31*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*a*d)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3795
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]
Rule 4001
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && !LtQ[m, -2^(-1)]
Rule 4010
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && !LtQ[m, -1]
Rule 4021
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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b
^2, 0] && GtQ[n, 1]
Rule 4088
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[(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 + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, 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 {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^3(c+d x) \left (\frac {1}{2} a (7 A+6 C)+\frac {1}{2} a (7 B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{7 a}\\ &=\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {4 \int \frac {\sec ^2(c+d x) \left (a^2 (7 B-C)+\frac {1}{4} a^2 (35 A-7 B+31 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A-7 B+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{8} a^3 (35 A-7 B+31 C)-\frac {1}{4} a^3 (35 A-49 B+37 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=-\frac {4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A-7 B+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(A-B+C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=-\frac {4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A-7 B+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}-\frac {(2 (A-B+C)) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {2} (A-B+C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (35 A-7 B+31 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 a d}\\ \end {align*}
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Mathematica [C] time = 11.38, size = 2322, normalized size = 11.16 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2*S
in[(c + d*x)/2]^2]*(-1/3*(A*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^(7/2) + (2*B*Sin[(c + d*x)/2])/(7*(1
- 2*Sin[(c + d*x)/2]^2)^(7/2)) + ((A - B + C)*Csc[(c + d*x)/2]^9*(363825*Sin[(c + d*x)/2]^2 - 4729725*Sin[(c +
d*x)/2]^4 + 26785605*Sin[(c + d*x)/2]^6 - 86790165*Sin[(c + d*x)/2]^8 + 177677808*Sin[(c + d*x)/2]^10 - 23928
3044*Sin[(c + d*x)/2]^12 + 52080*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^
2)]*Sin[(c + d*x)/2]^12 + 560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 +
2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12 + 213120160*Sin[(c + d*x)/2]^14 - 168280*Hypergeometric2F1[2, 11/2,
13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 2240*HypergeometricPFQ[{2, 2, 2,
2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 121497024*Sin
[(c + d*x)/2]^16 + 212520*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin
[(c + d*x)/2]^16 + 3360*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[
(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^16 + 40125184*Sin[(c + d*x)/2]^18 - 124320*Hypergeometric2F1[2, 11/2, 13/2,
Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^18 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2
}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^18 - 5840384*Sin[(c + d*x
)/2]^20 + 28000*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)
/2]^20 + 560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2
]^2)]*Sin[(c + d*x)/2]^20 + 363825*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sqrt[Sin[(c +
d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 5336100*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]
]*Sin[(c + d*x)/2]^2*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 34636140*ArcTanh[Sqrt[Sin[(c + d*x
)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^4*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] -
131060160*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^6*Sqrt[Sin[(c + d*x)
/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 320535600*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*S
in[(c + d*x)/2]^8*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 530671680*ArcTanh[Sqrt[Sin[(c + d*x)/
2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^10*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] +
604296000*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^12*Sqrt[Sin[(c + d*x)
/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 468948480*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*S
in[(c + d*x)/2]^14*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 237726720*ArcTanh[Sqrt[Sin[(c + d*x)
/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^16*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] -
70963200*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^18*Sqrt[Sin[(c + d*x)
/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 9461760*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin
[(c + d*x)/2]^20*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 1120*Cos[(c + d*x)/2]^6*Hypergeometric
PFQ[{2, 2, 2, 11/2}, {1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12*(-6 + 5
*Sin[(c + d*x)/2]^2) + 280*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 11/2}, {1, 13/2}, Sin[(c + d*x)/2]^2/(-
1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12*(103 - 164*Sin[(c + d*x)/2]^2 + 70*Sin[(c + d*x)/2]^4)))/(40425
*(1 - 2*Sin[(c + d*x)/2]^2)^(9/2)*(-1 + 2*Sin[(c + d*x)/2]^2)) + (4*B*((3*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*
x)/2]^2)^(5/2) + 4*(Sin[(c + d*x)/2]/(1 - 2*Sin[(c + d*x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c
+ d*x)/2]^2])))/35 + (A*((5*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^(7/2) + 2*((3*Sin[(c + d*x)/2])/(1 -
2*Sin[(c + d*x)/2]^2)^(5/2) + 4*(Sin[(c + d*x)/2]/(1 - 2*Sin[(c + d*x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2])/Sqr
t[1 - 2*Sin[(c + d*x)/2]^2]))))/105))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*
Sqrt[a*(1 + Sec[c + d*x])])
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fricas [A] time = 0.52, size = 448, normalized size = 2.15 \[ \left [\frac {105 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (35 \, A - 91 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A - 7 \, B + 31 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (7 \, B - C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left ({\left (35 \, A - 91 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (35 \, A - 7 \, B + 31 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (7 \, B - C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{105 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[1/210*(105*sqrt(2)*((A - B + C)*a*cos(d*x + c)^4 + (A - B + C)*a*cos(d*x + c)^3)*sqrt(-1/a)*log(-(2*sqrt(2)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c
) + 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((35*A - 91*B + 43*C)*cos(d*x + c)^3 - (35*A - 7*B + 31*C)*c
os(d*x + c)^2 - 3*(7*B - C)*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*co
s(d*x + c)^4 + a*d*cos(d*x + c)^3), -1/105*(2*((35*A - 91*B + 43*C)*cos(d*x + c)^3 - (35*A - 7*B + 31*C)*cos(d
*x + c)^2 - 3*(7*B - C)*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 105*sqrt(2
)*((A - B + C)*a*cos(d*x + c)^4 + (A - B + C)*a*cos(d*x + c)^3)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)]
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giac [A] time = 2.59, size = 305, normalized size = 1.47 \[ \frac {\frac {105 \, \sqrt {2} {\left (A - B + C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {2 \, {\left (\frac {105 \, \sqrt {2} B a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + {\left ({\left (\frac {\sqrt {2} {\left (70 \, A a^{3} - 119 \, B a^{3} + 92 \, C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {7 \, \sqrt {2} {\left (20 \, A a^{3} - 37 \, B a^{3} + 16 \, C a^{3}\right )}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {35 \, \sqrt {2} {\left (2 \, A a^{3} - 7 \, B a^{3} + 4 \, C a^{3}\right )}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{105 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
1/105*(105*sqrt(2)*(A - B + C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/
(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 2*(105*sqrt(2)*B*a^3/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + ((sqrt(2)*
(70*A*a^3 - 119*B*a^3 + 92*C*a^3)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 7*sqrt(2)*(20*A*a^3
- 37*B*a^3 + 16*C*a^3)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(2*A*a^3 - 7*B*a^
3 + 4*C*a^3)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1
/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
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maple [B] time = 2.04, size = 1144, normalized size = 5.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)
[Out]
1/840/d*(-105*A*cos(d*x+c)^3*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(
d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+105*B*sin(d*x+c)*cos(d*x+c)^3*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-105*C*sin(d*x+c)*cos(d*x+c)^3*
ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
7/2)-315*A*cos(d*x+c)^2*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c
))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+315*B*sin(d*x+c)*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-315*C*sin(d*x+c)*cos(d*x+c)^2*ln(((
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-
315*A*cos(d*x+c)*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+315*B*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-315*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-105*A*ln(((
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*
sin(d*x+c)+105*B*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(7/2)*sin(d*x+c)-105*C*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d
*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+560*A*cos(d*x+c)^4-1456*B*cos(d*x+c)^4+688*C*cos(d*x+c)
^4-1120*A*cos(d*x+c)^3+1568*B*cos(d*x+c)^3-1184*C*cos(d*x+c)^3+560*A*cos(d*x+c)^2-448*B*cos(d*x+c)^2+544*C*cos
(d*x+c)^2+336*B*cos(d*x+c)-288*C*cos(d*x+c)+240*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3/sin(d*x+c)
/a
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(1/2)),x)
[Out]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(1/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)
[Out]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/sqrt(a*(sec(c + d*x) + 1)), x)
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