3.1268 \(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=253 \[ \frac {a^{5/2} (40 A+38 B+25 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (24 A+42 B+31 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d \sqrt {\cos (c+d x)}}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\cos (c+d x)}} \]

[Out]

1/12*a*(6*B+5*C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+1/3*C*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/
d/cos(d*x+c)^(1/2)+1/8*a^(5/2)*(40*A+38*B+25*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+1/24*a^3*(24*A-54*B-49*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/24*a
^2*(24*A+42*B+31*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.87, 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, 4015, 3801, 215} \[ \frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (24 A+42 B+31 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{24 d \sqrt {\cos (c+d x)}}+\frac {a^{5/2} (40 A+38 B+25 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a (6 B+5 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d \sqrt {\cos (c+d x)}}+\frac {C \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(40*A + 38*B + 25*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt
[Sec[c + d*x]])/(8*d) + (a^3*(24*A - 54*B - 49*C)*Sin[c + d*x])/(24*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*
x]]) + (a^2*(24*A + 42*B + 31*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(24*d*Sqrt[Cos[c + d*x]]) + (a*(6*B +
5*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d*Sqrt[Cos[c + d*x]]) + (C*(a + a*Sec[c + d*x])^(5/2)*Sin[c
+ d*x])/(3*d*Sqrt[Cos[c + d*x]])

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 4015

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, 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 \sqrt {\cos (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 &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 {\sec (c+d x)}} \, dx\\ &=\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (6 A-C)+\frac {1}{2} a (6 B+5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (8 A-2 B-3 C)+\frac {1}{4} a^2 (24 A+42 B+31 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{6 a}\\ &=\frac {a^2 (24 A+42 B+31 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (24 A-54 B-49 C)+\frac {3}{8} a^3 (40 A+38 B+25 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{6 a}\\ &=\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (24 A+42 B+31 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {1}{16} \left (a^2 (40 A+38 B+25 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 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (24 A+42 B+31 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {\left (a^2 (40 A+38 B+25 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 )}{8 d}\\ &=\frac {a^{5/2} (40 A+38 B+25 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)}}{8 d}+\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (24 A+42 B+31 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d \sqrt {\cos (c+d x)}}+\frac {a (6 B+5 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d \sqrt {\cos (c+d x)}}+\frac {C (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.51, size = 157, normalized size = 0.62 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) (4 (18 A+6 B+17 C) \cos (c+d x)+3 (8 A+22 B+25 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+24 A+66 B+91 C)+3 \sqrt {2} (40 A+38 B+25 C) \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{48 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(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]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(40*A + 38*B + 25*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)
/2]]*Cos[c + d*x]^3 + (24*A + 66*B + 91*C + 4*(18*A + 6*B + 17*C)*Cos[c + d*x] + 3*(8*A + 22*B + 25*C)*Cos[2*(
c + d*x)] + 24*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d*Cos[c + d*x]^(5/2))

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fricas [A]  time = 0.81, size = 515, normalized size = 2.04 \[ \left [\frac {4 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\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 )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, 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 (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\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 )}{48 \, {\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(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/96*(4*(48*A*a^2*cos(d*x + c)^3 + 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c)
+ 8*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((40*A + 38*B + 25*C)*
a^2*cos(d*x + c)^4 + (40*A + 38*B + 25*C)*a^2*cos(d*x + c)^3)*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)^4 + d*cos(d*x + c)^3), 1/48*(2*(48*A*a^2*cos(d*x + c)^3
+ 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c)^2 + 2*(6*B + 17*C)*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c)^4 + (40*A + 38*B
+ 25*C)*a^2*cos(d*x + c)^3)*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)^4 + d*cos(d*x + c)^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 = 1.88, size = 567, normalized size = 2.24 \[ -\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 (96 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 )}}+120 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 ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-120 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 ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+114 B \left (\cos ^{3}\left (d x +c \right )\right ) \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 )-114 B \left (\cos ^{3}\left (d x +c \right )\right ) \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 )+75 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 ^{3}\left (d x +c \right )\right )-75 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 ^{3}\left (d x +c \right )\right )+48 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 )}}+132 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 )}}+150 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 )+24 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+68 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{\frac {5}{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(cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-1/48/d*a^2*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(96*A*cos(d*x+c)^3*sin(d*x+c)*(-2/(1+cos(d*x+c
)))^(1/2)+120*A*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)^3-1
20*A*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)^3+114*B*cos(d*
x+c)^3*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))-114*B*cos(d*x+c)^3*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))+75*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)^3-75*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)^3+48*A*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+132
*B*sin(d*x+c)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+150*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^2+2
4*B*sin(d*x+c)*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+68*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+16*C*
(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))/cos(d*x+c)^(5/2)/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(cos(d*x+c)^(1/2)*(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 = -0.00 \[ \int \sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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