3.9.49 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [849]

Optimal. Leaf size=509 \[ -\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-16 a^5 C+28 a^3 b^2 C-8 a b^4 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}-\frac {2 \left (a^3 b (8 B-12 C)-9 a b^3 (B-C)-b^4 (3 B-C)-16 a^4 C+2 a^2 b^2 (3 B+8 C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]

[Out]

-2/3*(8*B*a^4*b-15*B*a^2*b^3+3*B*b^5-16*C*a^5+28*C*a^3*b^2-8*C*a*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1
/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^5/
(a+b)^(3/2)/d-2/3*(a^3*b*(8*B-12*C)-9*a*b^3*(B-C)-b^4*(3*B-C)-16*a^4*C+2*a^2*b^2*(3*B+8*C))*cot(d*x+c)*Ellipti
cF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(
a-b))^(1/2)/b^4/(a^2-b^2)/d/(a+b)^(1/2)+2/3*a*(B*b-C*a)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))
^(3/2)-2/3*a^2*(3*B*a^2*b-7*B*b^3-6*C*a^3+10*C*a*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)-2/3*
(B*a*b-2*C*a^2+C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d

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Rubi [A]
time = 1.09, antiderivative size = 509, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4157, 4114, 4175, 4167, 4090, 3917, 4089} \begin {gather*} \frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-16 a^4 C+a^3 b (8 B-12 C)+2 a^2 b^2 (3 B+8 C)-9 a b^3 (B-C)-b^4 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {2 \left (-16 a^5 C+8 a^4 b B+28 a^3 b^2 C-15 a^2 b^3 B-8 a b^4 C+3 b^5 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^5 d (a-b) (a+b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 16*a^5*C + 28*a^3*b^2*C - 8*a*b^4*C)*Cot[c + d*x]*EllipticE[ArcSin[S
qrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec
[c + d*x]))/(a - b))])/(3*(a - b)*b^5*(a + b)^(3/2)*d) - (2*(a^3*b*(8*B - 12*C) - 9*a*b^3*(B - C) - b^4*(3*B -
C) - 16*a^4*C + 2*a^2*b^2*(3*B + 8*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (
a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4*Sqrt[a +
b]*(a^2 - b^2)*d) + (2*a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2
)) - (2*a^2*(3*a^2*b*B - 7*b^3*B - 6*a^3*C + 10*a*b^2*C)*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c
+ d*x]]) - (2*(a*b*B - 2*a^2*C + b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(a^2 - b^2)*d)

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

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

Rule 4114

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

Rule 4157

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

Rule 4167

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4175

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e
+ f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {\sec ^2(c+d x) \left (2 a (b B-a C)-\frac {3}{2} b (b B-a C) \sec (c+d x)-\frac {3}{2} \left (a b B-2 a^2 C+b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\sec (c+d x) \left (\frac {1}{4} a b \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )+\frac {1}{4} \left (6 a^4 b B-13 a^2 b^3 B+3 b^5 B-12 a^5 C+22 a^3 b^2 C-6 a b^4 C\right ) \sec (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (a b B-2 a^2 C+b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {3}{8} b^2 \left (2 a^3 b B-6 a b^3 B-4 a^4 C+7 a^2 b^2 C+b^4 C\right )+\frac {3}{8} b \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-16 a^5 C+28 a^3 b^2 C-8 a b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-16 a^5 C+28 a^3 b^2 C-8 a b^4 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}-\frac {\left (a^3 b (8 B-12 C)-9 a b^3 (B-C)-b^4 (3 B-C)-16 a^4 C+2 a^2 b^2 (3 B+8 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b^3 (a+b)^2}\\ &=-\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-16 a^5 C+28 a^3 b^2 C-8 a b^4 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}-\frac {2 \left (a^3 b (8 B-12 C)-9 a b^3 (B-C)-b^4 (3 B-C)-16 a^4 C+2 a^2 b^2 (3 B+8 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}+\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4342\) vs. \(2(509)=1018\).
time = 27.01, size = 4342, normalized size = 8.53 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 16*a^5*C + 28*a^3*b^2*C - 8*a
*b^4*C)*Sin[c + d*x])/(3*b^4*(-a^2 + b^2)^2) + (2*(a^2*b*B*Sin[c + d*x] - a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 +
b^2)*(b + a*Cos[c + d*x])^2) + (2*(-4*a^4*b*B*Sin[c + d*x] + 8*a^2*b^3*B*Sin[c + d*x] + 7*a^5*C*Sin[c + d*x] -
11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (2*C*Tan[c + d*x])/(3*b^3)))/(d*(a
+ b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*((5*a^2*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[
Sec[c + d*x]]) - (8*a^4*B)/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (b^2*B)/((-a^2
+ b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a^5*C)/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d
*x]]*Sqrt[Sec[c + d*x]]) - (28*a^3*C)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a*
b*C)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^5*B*Sqrt[Sec[c + d*x]])/(3*b^3*(-a^
2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (17*a^3*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*
x]]) - (3*a*b*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (5*a^2*C*Sqrt[Sec[c + d*x]])/(
(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (16*a^6*C*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[
c + d*x]]) - (32*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (b^2*C*Sqrt[Sec[c
+ d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^5*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^3*(
-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (5*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(b*(-a^2 + b^2)^2*Sqrt
[b + a*Cos[c + d*x]]) - (a*b*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]])
+ (8*a^2*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (16*a^6*C*Cos[2*
(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (28*a^4*C*Cos[2*(c + d*x)]*Sq
rt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]))*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/2]^2*
Sec[c + d*x]]*(2*(a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Sqrt[Cos[
c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c +
d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-16*a^4*C - 9*a*b^3*(B + C) + b^4*(3*B + C) + 4*a^3*b*(2*B + 3*C) +
2*a^2*b^2*(-3*B + 8*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*
C - 28*a^3*b^2*C + 8*a*b^4*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(
a^2 - b^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2)*((a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*S
in[c + d*x]*(2*(a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Sqrt[Cos[c
+ d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d
*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-16*a^4*C - 9*a*b^3*(B + C) + b^4*(3*B + C) + 4*a^3*b*(2*B + 3*C) + 2
*a^2*b^2*(-3*B + 8*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d
*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C
- 28*a^3*b^2*C + 8*a*b^4*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(a^
2 - b^2)^2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) - (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(
c + d*x)/2]*(2*(a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Sqrt[Cos[c
+ d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d
*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(-16*a^4*C - 9*a*b^3*(B + C) + b^4*(3*B + C) + 4*a^3*b*(2*B + 3*C) + 2
*a^2*b^2*(-3*B + 8*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d
*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C
- 28*a^3*b^2*C + 8*a*b^4*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(a^
2 - b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((-8
*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec
[(c + d*x)/2]^4)/2 + ((a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*b^4*C)*Sqrt
[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*((Cos
[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c
+ d*x])] + (b*(a + b)*(-16*a^4*C - 9*a*b^3*(B + C) + b^4*(3*B + C) + 4*a^3*b*(2*B + 3*C) + 2*a^2*b^2*(-3*B + 8
*C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Co...

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(8043\) vs. \(2(475)=950\).
time = 0.76, size = 8044, normalized size = 15.80

method	result	size
default	\(\text {Expression too large to display}\)	\(8044\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a)/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x
+ c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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