3.1334 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.94, 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 = {4221, 3045, 2976, 2981, 2774, 216} \[ \frac {a^{5/2} (304 A+200 B+163 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (16 A+24 B+17 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{32 d \sqrt {\sec (c+d x)}}+\frac {a (8 B+5 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{24 d \sqrt {\sec (c+d x)}}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 216

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

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 3045

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*
sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 4221

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

Rubi steps

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

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Mathematica [A]  time = 1.46, size = 166, normalized size = 0.66 \[ \frac {a^2 \sqrt {\cos (c+d x)} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {a (\cos (c+d x)+1)} \left (3 \sqrt {2} (304 A+200 B+163 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} ((96 A+272 B+362 C) \cos (c+d x)+528 A+4 (8 B+23 C) \cos (2 (c+d x))+632 B+12 C \cos (3 (c+d x))+581 C)\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(3*Sqrt[2]*(304*A + 200
*B + 163*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(528*A + 632*B + 581*C + (96*A + 272*B + 3
62*C)*Cos[c + d*x] + 4*(8*B + 23*C)*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(384*d)

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fricas [A]  time = 1.02, size = 195, normalized size = 0.77 \[ -\frac {3 \, {\left ({\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (48 \, C a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, B + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 136 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (176 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/192*(3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c) + (304*A + 200*B + 163*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d
*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (48*C*a^2*cos(d*x + c)^4 + 8*(8*B + 23*C)*a^2*cos(d*
x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c))*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

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giac [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*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.56, size = 477, normalized size = 1.89 \[ -\frac {\left (48 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+64 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+184 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+96 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+272 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+326 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+528 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+600 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+489 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+912 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+600 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+489 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right ) a^{2}}{192 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192/d*(48*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+64*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)+184*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+96*A*sin(d*x+c)*cos(
d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+272*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+326*C*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+528*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+600*
B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+489*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+912*A*arctan
(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+600*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)/cos(d*x+c))+489*C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*(1/cos(d*x+c))^(1/2)*
(a*(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2*(cos(d*x+c)^2-1)*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*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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 \[ \int \sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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