3.1343 \(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=356 \[ \frac {2 \sqrt {\cos (c+d x)} \left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \left (-5 a^3 B-3 a^2 b (A+5 C)+5 a b^2 B+3 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac {2 b^2 C \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
[Out]
2/5*A*cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d-2/15*(3*A*b^3-5*a^3*B+5*a*b^2*B-3*a^2*b*(A+5*C))*(c
os(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(
d*x+c))/(a+b))^(1/2)/a/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2*b^2*C*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*
d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c
)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/15*(3*A*b+5*B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d+2/15*(3
*A*b^2+20*a*b*B+3*a^2*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),
2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
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Rubi [A] time = 1.40, antiderivative size = 356, normalized size of antiderivative = 1.00,
number of steps used = 14, number of rules used = 13, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules
used = {4265, 4094, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ -\frac {2 \left (-3 a^2 b (A+5 C)-5 a^3 B+5 a b^2 B+3 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac {2 b^2 C \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(3*A*b^3 - 5*a^3*B + 5*a*b^2*B - 3*a^2*b*(A + 5*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)
/2, (2*a)/(a + b)])/(15*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b^2*C*Sqrt[(b + a*Cos[c + d*x])/
(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*A*
b^2 + 20*a*b*B + 3*a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c
+ d*x]])/(15*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*(3*A*b + 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c
+ d*x]]*Sin[c + d*x])/(15*d) + (2*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 2807
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Rule 3856
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 3858
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 3859
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4035
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Rule 4094
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[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Rule 4108
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 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 \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \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+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {1}{2} (3 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)+\frac {5}{2} b C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (12 a A b+5 a^2 B+15 b^2 B+30 a b C\right ) \sec (c+d x)+\frac {15}{4} b^2 C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right )+\frac {1}{4} \left (12 a A b+5 a^2 B+15 b^2 B+30 a b C\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx+\left (b^2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (\left (3 A b^3-5 a^3 B+5 a b^2 B-3 a^2 b (A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a}+\frac {\left (\left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{15 a}+\frac {\left (b^2 C \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ &=\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (\left (3 A b^3-5 a^3 B+5 a b^2 B-3 a^2 b (A+5 C)\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{15 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (b^2 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{15 a \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 b^2 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (\left (3 A b^3-5 a^3 B+5 a b^2 B-3 a^2 b (A+5 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{15 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{15 a \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=-\frac {2 \left (3 A b^3-5 a^3 B+5 a b^2 B-3 a^2 b (A+5 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 b^2 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 (3 A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}
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Mathematica [C] time = 35.15, size = 56321, normalized size = 158.21 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
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fricas [F] time = 2.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} + {\left (C a + B b\right )} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )^{2} + {\left (B a + A b\right )} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*b*cos(d*x + c)^2*sec(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2*sec(d*x + c)^2 + A*a*cos(d*x + c)^2 +
(B*a + A*b)*cos(d*x + c)^2*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2)*(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)*(b*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2), x)
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maple [C] time = 2.15, size = 2220, normalized size = 6.24 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-2/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^3*(9*A*sin(d*x+c
)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)+6*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)
))^(3/2)+9*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*a^3-9*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3+3*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3-5*B*EllipticF((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^3-15*C*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-
b))^(1/2))*a^3+15*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2
)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3+5*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)+20*B*
sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+15*C*((a-b)/(a+b))^(1/2)*a^2*b*sin(d*x+c)*(1/(1+
cos(d*x+c)))^(3/2)+3*A*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+9*A*cos(d*x+c)
*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+15*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*sin(d*x+c
)*(1/(1+cos(d*x+c)))^(3/2)+5*B*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(3/2)*a^3+5*B*co
s(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(3/2)*a^3+3*A*cos(d*x+c)^3*sin(d*x+c)*((a-b)/(a+b))
^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+3*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(3/2)+9*A*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b)
)^(1/2))*a^2*b-3*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)
/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2-20*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b+20*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+20*B*EllipticF((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*a^2*b+15*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d
*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-30*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))
*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-12*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)
/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^2*b+3*A*EllipticF((-1+cos(d*
x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^2
-15*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*a*b^2+15*C*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^2-30*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b^2+9*A*cos(d*x+c)^2*
sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)+9*A*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^
2*b*(1/(1+cos(d*x+c)))^(3/2)+9*A*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+25*B
*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2))/a/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+
c))/sin(d*x+c)^6/(1/(1+cos(d*x+c)))^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(3/2)*(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)**(5/2)*(a+b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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