3.1269 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\)

Optimal. Leaf size=253 \[ \frac {a^{5/2} (304 A+200 B+163 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (16 A+24 B+17 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (8 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/24*a*(8*B+5*C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+1/4*C*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/
d/cos(d*x+c)^(3/2)+1/64*a^(5/2)*(304*A+200*B+163*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x
+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/192*a^3*(432*A+392*B+299*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2
)+1/32*a^2*(16*A+24*B+17*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)

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Rubi [A]  time = 0.89, antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4265, 4088, 4018, 4016, 3801, 215} \[ \frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (16 A+24 B+17 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a^{5/2} (304 A+200 B+163 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a (8 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d \cos ^{\frac {3}{2}}(c+d x)} \]

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))/Sqrt[Cos[c + d*x]],x]

[Out]

(a^(5/2)*(304*A + 200*B + 163*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]])/(64*d) + (a^3*(432*A + 392*B + 299*C)*Sin[c + d*x])/(192*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Se
c[c + d*x]]) + (a^2*(16*A + 24*B + 17*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(32*d*Cos[c + d*x]^(3/2)) + (a
*(8*B + 5*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(24*d*Cos[c + d*x]^(3/2)) + (C*(a + a*Sec[c + d*x])^(5/2
)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2))

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 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))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (8 A+C)+\frac {1}{2} a (8 B+5 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {a (8 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (48 A+8 B+11 C)+\frac {3}{4} a^2 (16 A+24 B+17 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a^2 (16 A+24 B+17 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (8 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (240 A+104 B+95 C)+\frac {1}{8} a^3 (432 A+392 B+299 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (16 A+24 B+17 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (8 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{128} \left (a^2 (304 A+200 B+163 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^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (16 A+24 B+17 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (8 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (a^2 (304 A+200 B+163 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 )}{64 d}\\ &=\frac {a^{5/2} (304 A+200 B+163 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)}}{64 d}+\frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (16 A+24 B+17 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {a (8 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 3.91, size = 178, normalized size = 0.70 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) ((1584 A+2056 B+2203 C) \cos (c+d x)+4 (48 A+136 B+163 C) \cos (2 (c+d x))+528 A \cos (3 (c+d x))+192 A+600 B \cos (3 (c+d x))+544 B+489 C \cos (3 (c+d x))+844 C)+6 \sqrt {2} (304 A+200 B+163 C) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{768 d \cos ^{\frac {7}{2}}(c+d x)} \]

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))/Sqrt[Cos[c + d*x]],x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(304*A + 200*B + 163*C)*ArcTanh[Sqrt[2]*Sin[(c + d
*x)/2]]*Cos[c + d*x]^4 + (192*A + 544*B + 844*C + (1584*A + 2056*B + 2203*C)*Cos[c + d*x] + 4*(48*A + 136*B +
163*C)*Cos[2*(c + d*x)] + 528*A*Cos[3*(c + d*x)] + 600*B*Cos[3*(c + d*x)] + 489*C*Cos[3*(c + d*x)])*Sin[(c + d
*x)/2]))/(768*d*Cos[c + d*x]^(7/2))

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fricas [A]  time = 0.95, size = 533, normalized size = 2.11 \[ \left [\frac {4 \, {\left (3 \, {\left (176 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 136 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 23 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\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 ) + 3 \, {\left ({\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\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 )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (3 \, {\left (176 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 136 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, B + 23 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\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 ) + 3 \, {\left ({\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\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 )}{384 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\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)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(4*(3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 8*(8*B
+ 23*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)
+ 3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c)^5 + (304*A + 200*B + 163*C)*a^2*cos(d*x + c)^4)*sqrt(a)*log((a*c
os(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)^5 + d*cos(d*x + c)^4),
1/384*(2*(3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 8*(8*B
+ 23*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) +
3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c)^5 + (304*A + 200*B + 163*C)*a^2*cos(d*x + c)^4)*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)^5 + d*cos(d*x + c)^4)]

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

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)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.35, size = 629, normalized size = 2.49 \[ -\frac {a^{2} \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 (912 A \left (\cos ^{4}\left (d x +c \right )\right ) \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 ) \sqrt {2}-912 A \left (\cos ^{4}\left (d x +c \right )\right ) \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 ) \sqrt {2}+600 B \left (\cos ^{4}\left (d x +c \right )\right ) \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 ) \sqrt {2}-600 B \left (\cos ^{4}\left (d x +c \right )\right ) \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 ) \sqrt {2}+489 C \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 ) \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )-489 C \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 ) \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )+1056 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 )}}+1200 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 )}}+978 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 )+192 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 )}}+544 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 )}}+652 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 )+128 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+368 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+96 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right )}{384 d \cos \left (d x +c \right )^{\frac {7}{2}} \sin \left (d x +c \right )^{2} \sqrt {-\frac {2}{1+\cos \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)/cos(d*x+c)^(1/2),x)

[Out]

-1/384/d*a^2*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(912*A*cos(d*x+c)^4*arctan(1/4*(-2/(1+cos(d*x
+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)-912*A*cos(d*x+c)^4*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)
*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)+600*B*cos(d*x+c)^4*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c
)+1+sin(d*x+c))*2^(1/2))*2^(1/2)-600*B*cos(d*x+c)^4*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x
+c))*2^(1/2))*2^(1/2)+489*C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)*co
s(d*x+c)^4-489*C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)^4+
1056*A*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+1200*B*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1
/2)+978*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^3+192*A*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^
(1/2)+544*B*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+652*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d
*x+c)^2+128*B*sin(d*x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+368*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d
*x+c)+96*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))/cos(d*x+c)^(7/2)/sin(d*x+c)^2/(-2/(1+cos(d*x+c)))^(1/2)

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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)/cos(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 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \]

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))/cos(c + d*x)^(1/2),x)

[Out]

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

[Out]

Timed out

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