3.1261 \(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=303 \[ \frac {a^{3/2} (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a (10 B+3 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

1/5*C*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(7/2)+1/128*a^(3/2)*(176*A+150*B+133*C)*arcsinh(a^(1/2)*t
an(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/240*a^2*(80*A+90*B+67*C)*sin(d*x+c)/d/
cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2)+1/192*a^2*(176*A+150*B+133*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d
*x+c))^(1/2)+1/128*a^2*(176*A+150*B+133*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/40*a*(10*B+3
*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)

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Rubi [A]  time = 0.87, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4265, 4088, 4018, 4016, 3803, 3801, 215} \[ \frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^{3/2} (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a (10 B+3 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(a^(3/2)*(176*A + 150*B + 133*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*S
qrt[Sec[c + d*x]])/(128*d) + (a^2*(80*A + 90*B + 67*C)*Sin[c + d*x])/(240*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[
c + d*x]]) + (a^2*(176*A + 150*B + 133*C)*Sin[c + d*x])/(192*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) +
(a^2*(176*A + 150*B + 133*C)*Sin[c + d*x])/(128*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(10*B + 3*
C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(40*d*Cos[c + d*x]^(7/2)) + (C*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*
x])/(5*d*Cos[c + d*x]^(7/2))

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 3803

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*d
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(2*a*d*(n - 1))/(b*(
2*n - 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 4016

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

Rule 4018

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] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[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]

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {5}{2} a (2 A+C)+\frac {1}{2} a (10 B+3 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {5}{4} a^2 (16 A+10 B+11 C)+\frac {1}{4} a^2 (80 A+90 B+67 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{96} \left (a (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{128} \left (a (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{256} \left (a (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {\left (a (176 A+150 B+133 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}\\ &=\frac {a^{3/2} (176 A+150 B+133 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{128 d}+\frac {a^2 (80 A+90 B+67 C) \sin (c+d x)}{240 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{192 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (176 A+150 B+133 C) \sin (c+d x)}{128 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (10 B+3 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 5.81, size = 210, normalized size = 0.69 \[ \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) (12 (880 A+1070 B+1273 C) \cos (c+d x)+4 (3280 A+3450 B+3059 C) \cos (2 (c+d x))+3520 A \cos (3 (c+d x))+2640 A \cos (4 (c+d x))+10480 A+3000 B \cos (3 (c+d x))+2250 B \cos (4 (c+d x))+11550 B+2660 C \cos (3 (c+d x))+1995 C \cos (4 (c+d x))+13313 C)+60 \sqrt {2} (176 A+150 B+133 C) \cos ^5(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{15360 d \cos ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(5/2),x]

[Out]

(a*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(60*Sqrt[2]*(176*A + 150*B + 133*C)*ArcTanh[Sqrt[2]*Sin[(c + d*
x)/2]]*Cos[c + d*x]^5 + (10480*A + 11550*B + 13313*C + 12*(880*A + 1070*B + 1273*C)*Cos[c + d*x] + 4*(3280*A +
3450*B + 3059*C)*Cos[2*(c + d*x)] + 3520*A*Cos[3*(c + d*x)] + 3000*B*Cos[3*(c + d*x)] + 2660*C*Cos[3*(c + d*x
)] + 2640*A*Cos[4*(c + d*x)] + 2250*B*Cos[4*(c + d*x)] + 1995*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]))/(15360*d*
Cos[c + d*x]^(9/2))

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fricas [A]  time = 0.88, size = 551, normalized size = 1.82 \[ \left [\frac {4 \, {\left (15 \, {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{4} + 10 \, {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 19 \, C\right )} a \cos \left (d x + c\right ) + 384 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 15 \, {\left ({\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{6} + {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}, \frac {2 \, {\left (15 \, {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{4} + 10 \, {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (80 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{2} + 48 \, {\left (10 \, B + 19 \, C\right )} a \cos \left (d x + c\right ) + 384 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 15 \, {\left ({\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{6} + {\left (176 \, A + 150 \, B + 133 \, C\right )} a \cos \left (d x + c\right )^{5}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{3840 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/7680*(4*(15*(176*A + 150*B + 133*C)*a*cos(d*x + c)^4 + 10*(176*A + 150*B + 133*C)*a*cos(d*x + c)^3 + 8*(80*
A + 150*B + 133*C)*a*cos(d*x + c)^2 + 48*(10*B + 19*C)*a*cos(d*x + c) + 384*C*a)*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((176*A + 150*B + 133*C)*a*cos(d*x + c)^6 + (176*A + 150*B + 1
33*C)*a*cos(d*x + c)^5)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos
(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))
/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5), 1/3840*(2*(15*(176*A + 150*B + 133*C)*a*cos(d*x + c)^4 + 10*(176*A + 1
50*B + 133*C)*a*cos(d*x + c)^3 + 8*(80*A + 150*B + 133*C)*a*cos(d*x + c)^2 + 48*(10*B + 19*C)*a*cos(d*x + c) +
384*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 15*((176*A + 150*B + 133*C
)*a*cos(d*x + c)^6 + (176*A + 150*B + 133*C)*a*cos(d*x + c)^5)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c
)^6 + d*cos(d*x + c)^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)

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maple [B]  time = 2.29, size = 720, normalized size = 2.38 \[ \frac {a \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (2640 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}-2640 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}+2250 B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}-2250 B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}+1995 C \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right )-1995 C \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{5}\left (d x +c \right )\right )-5280 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4500 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3990 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3520 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-3000 B \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-2660 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-1280 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-2400 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-2128 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-960 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-1824 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )-768 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right )}{3840 d \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x)

[Out]

1/3840/d*a*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(2640*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(c
os(d*x+c)+1-sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)-2640*A*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+
1+sin(d*x+c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)+2250*B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+
c))*2^(1/2))*cos(d*x+c)^5*2^(1/2)-2250*B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2
))*cos(d*x+c)^5*2^(1/2)+1995*C*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))
*cos(d*x+c)^5-1995*C*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*cos(d*x+c
)^5-5280*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)-4500*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin
(d*x+c)-3990*C*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)-3520*A*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x
+c)))^(1/2)-3000*B*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-2660*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)*cos(d*x+c)^3-1280*A*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-2400*B*sin(d*x+c)*cos(d*x+c)^2*(-2/(1
+cos(d*x+c)))^(1/2)-2128*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^2-960*B*sin(d*x+c)*cos(d*x+c)*(-2/(
1+cos(d*x+c)))^(1/2)-1824*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)-768*C*(-2/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c))/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^(9/2)

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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))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(5/2),x)

[Out]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(5/2), x)

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sympy [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))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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