3.503 \(\int (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=182 \[ \frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

[Out]

2*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/35*a*(7*B+5*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*
x+c)/d+2/7*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/105*a^3*(245*A+224*B+160*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(
1/2)+2/105*a^2*(35*A+56*B+40*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.32, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4054, 3917, 3915, 3774, 203, 3792} \[ \frac {2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (35 A+56 B+40 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{105 d}+\frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a (7 B+5 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{35 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(245*A + 224*B + 160*C)*Tan[c
+ d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(35*A + 56*B + 40*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])
/(105*d) + (2*a*(7*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*d) + (2*C*(a + a*Sec[c + d*x])^(5/2)*
Tan[c + d*x])/(7*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 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3792

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

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4054

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

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2 \int (a+a \sec (c+d x))^{5/2} \left (\frac {7 a A}{2}+\frac {1}{2} a (7 B+5 C) \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {4 \int (a+a \sec (c+d x))^{3/2} \left (\frac {35 a^2 A}{4}+\frac {1}{4} a^2 (35 A+56 B+40 C) \sec (c+d x)\right ) \, dx}{35 a}\\ &=\frac {2 a^2 (35 A+56 B+40 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {8 \int \sqrt {a+a \sec (c+d x)} \left (\frac {105 a^3 A}{8}+\frac {1}{8} a^3 (245 A+224 B+160 C) \sec (c+d x)\right ) \, dx}{105 a}\\ &=\frac {2 a^2 (35 A+56 B+40 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\left (a^2 A\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{105} \left (a^2 (245 A+224 B+160 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (35 A+56 B+40 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}-\frac {\left (2 a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (245 A+224 B+160 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (35 A+56 B+40 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 2.44, size = 170, normalized size = 0.93 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt {a (\sec (c+d x)+1)} \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) ((840 A+987 B+930 C) \cos (c+d x)+2 (35 A+98 B+115 C) \cos (2 (c+d x))+280 A \cos (3 (c+d x))+70 A+301 B \cos (3 (c+d x))+196 B+230 C \cos (3 (c+d x))+290 C)+420 \sqrt {2} A \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^3*Sqrt[a*(1 + Sec[c + d*x])]*(420*Sqrt[2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]
]*Cos[c + d*x]^(7/2) + 2*(70*A + 196*B + 290*C + (840*A + 987*B + 930*C)*Cos[c + d*x] + 2*(35*A + 98*B + 115*C
)*Cos[2*(c + d*x)] + 280*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)] + 230*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2
]))/(420*d)

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fricas [A]  time = 0.49, size = 430, normalized size = 2.36 \[ \left [\frac {105 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, -\frac {2 \, {\left (105 \, {\left (A a^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (280 \, A + 301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (35 \, A + 98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((
a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((28
0*A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*
x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
, -2/105*(105*(A*a^2*cos(d*x + c)^4 + A*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((280*A + 301*B + 230*C)*a^2*cos(d*x + c)^3 + (35*A + 98*B + 115*C
)*a^2*cos(d*x + c)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin
(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]

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giac [B]  time = 2.56, size = 419, normalized size = 2.30 \[ -\frac {\frac {105 \, A \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\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 )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\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 )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (315 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 420 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (875 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 980 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 700 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (805 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 784 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 560 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (245 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 224 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 160 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\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 )^{2}\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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/105*(105*A*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 -
4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqr
t(2)*abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + 2*(315*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 420*sqrt(2)*B*a^6*sgn(
cos(d*x + c)) + 420*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (875*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 980*sqrt(2)*B*a^6
*sgn(cos(d*x + c)) + 700*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (805*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 784*sqrt(2)*
B*a^6*sgn(cos(d*x + c)) + 560*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (245*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 224*sqr
t(2)*B*a^6*sgn(cos(d*x + c)) + 160*sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*
c)^2)*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 = 1.62, size = 476, normalized size = 2.62 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-105 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-315 A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-105 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sin \left (d x +c \right )+4480 A \left (\cos ^{4}\left (d x +c \right )\right )+4816 B \left (\cos ^{4}\left (d x +c \right )\right )+3680 C \left (\cos ^{4}\left (d x +c \right )\right )-3920 A \left (\cos ^{3}\left (d x +c \right )\right )-3248 B \left (\cos ^{3}\left (d x +c \right )\right )-1840 C \left (\cos ^{3}\left (d x +c \right )\right )-560 A \left (\cos ^{2}\left (d x +c \right )\right )-1232 B \left (\cos ^{2}\left (d x +c \right )\right )-880 C \left (\cos ^{2}\left (d x +c \right )\right )-336 B \cos \left (d x +c \right )-720 C \cos \left (d x +c \right )-240 C \right ) a^{2}}{840 d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/840/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-105*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*arctanh(1/2*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-315*A*2^(1/2)*sin(
d*x+c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(7/2)-315*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-105*A*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+4480*A*c
os(d*x+c)^4+4816*B*cos(d*x+c)^4+3680*C*cos(d*x+c)^4-3920*A*cos(d*x+c)^3-3248*B*cos(d*x+c)^3-1840*C*cos(d*x+c)^
3-560*A*cos(d*x+c)^2-1232*B*cos(d*x+c)^2-880*C*cos(d*x+c)^2-336*B*cos(d*x+c)-720*C*cos(d*x+c)-240*C)/cos(d*x+c
)^3/sin(d*x+c)*a^2

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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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(5/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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