∫ (a + b sec(c + dx))5∕2(A + B sec(c + dx)) dx [366]

3.4.66 \(\int (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [366]

Optimal. Leaf size=442 \[ -\frac {2 (a-b) \sqrt {a+b} \left (35 a A b+23 a^2 B+9 b^2 B\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {2 \sqrt {a+b} \left (a^2 b (45 A-23 B)-a b^2 (35 A-17 B)+b^3 (5 A-9 B)+15 a^3 B\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b (5 A b+8 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

-2/15*(a-b)*(35*A*a*b+23*B*a^2+9*B*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^

(1/2))*(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/d+2/15*(a^2*b*(45*A-23*B)-

a*b^2*(35*A-17*B)+b^3*(5*A-9*B)+15*a^3*B)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b)

)^(1/2))*(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/d-2*a^2*A*cot(d*x+c)*Ell

ipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(

1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+2/5*b*B*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/15*b*(5*A*b+8*B*a)*(a+b*s

ec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.42, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4003, 4141, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\frac {2 (a-b) \sqrt {a+b} \left (23 a^2 B+35 a A b+9 b^2 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 )}{15 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (15 a^3 B+a^2 b (45 A-23 B)-a b^2 (35 A-17 B)+b^3 (5 A-9 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}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}+\frac {2 b (8 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 b B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(35*a*A*b + 23*a^2*B + 9*b^2*B)*Cot[c + d*x]*EllipticE[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))])/

(15*b*d) + (2*Sqrt[a + b]*(a^2*b*(45*A - 23*B) - a*b^2*(35*A - 17*B) + b^3*(5*A - 9*B) + 15*a^3*B)*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))])/(15*b*d) - (2*a^2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a,

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))])/d + (2*b*(5*A*b + 8*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*b*B

*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b

*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b

*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

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 4003

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-b)

*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2

*c*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x],

x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4006

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In

t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,

b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4141

Int[((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/(f*(m + 1))), x] + Dist[1/(m + 1), I

nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +

a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]

Rule 4143

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_

.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, 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, C}, x] && NeQ[a^2 - b^2, 0

]

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \sec (c+d x)} \left (\frac {5 a^2 A}{2}+\frac {1}{2} \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sec (c+d x)+\frac {1}{2} b (5 A b+8 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {2 b (5 A b+8 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15 a^3 A}{4}+\frac {1}{4} \left (45 a^2 A b+5 A b^3+15 a^3 B+17 a b^2 B\right ) \sec (c+d x)+\frac {1}{4} b \left (35 a A b+23 a^2 B+9 b^2 B\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 b (5 A b+8 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15 a^3 A}{4}+\left (-\frac {1}{4} b \left (35 a A b+23 a^2 B+9 b^2 B\right )+\frac {1}{4} \left (45 a^2 A b+5 A b^3+15 a^3 B+17 a b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{15} \left (b \left (35 a A b+23 a^2 B+9 b^2 B\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a A b+23 a^2 B+9 b^2 B\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}}}{15 b d}+\frac {2 b (5 A b+8 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\left (a^3 A\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{15} \left (a^2 b (45 A-23 B)-a b^2 (35 A-17 B)+b^3 (5 A-9 B)+15 a^3 B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a A b+23 a^2 B+9 b^2 B\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}}}{15 b d}+\frac {2 \sqrt {a+b} \left (a^2 b (45 A-23 B)-a b^2 (35 A-17 B)+b^3 (5 A-9 B)+15 a^3 B\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}}}{15 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{a};\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}}}{d}+\frac {2 b (5 A b+8 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(7138\) vs. \(2(442)=884\).
time = 25.45, size = 7138, normalized size = 16.15 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3284\) vs. \(2(403)=806\).
time = 11.08, size = 3285, normalized size = 7.43


method	result	size

default	\(\text {Expression too large to display}\)	\(3285\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-5*A*cos(d*x+c)^3*b^3+23*B*cos(

d*x+c)^3*a^3-9*B*cos(d*x+c)^3*b^3+6*B*cos(d*x+c)^2*b^3-23*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))

)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)

)*a^2*b+3*B*b^3+15*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1

/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-30*A*cos(d*x+c)^2*(cos(d*x+c)/(1+

cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-

b)/(a+b))^(1/2))*sin(d*x+c)*a^3-15*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d

*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3+15*A*cos(d*x+c)^3

*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin

(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-30*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+

c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-1

5*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1

+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-5*B*cos(d*x+c)^3*a*b^2-17*B*cos(d*x+c)^2*sin(d*x+c

)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/si

n(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+23*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d

*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+9*B*cos(d*x

+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1

+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-45*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1

/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^

2*b-35*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/

2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+35*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1

+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/

(a+b))^(1/2))*a^2*b+35*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*

x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-23*B*cos(d*x+c)^3*sin(d*x+c

)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/si

n(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-17*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d

*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+23*B*cos(d*

x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-

1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+9*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1

/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*

b^2-45*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/

2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-35*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1

+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/

(a+b))^(1/2))*a*b^2+35*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*

x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+35*A*cos(d*x+c)^2*sin(d*x+c

)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/si

n(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+5*A*cos(d*x+c)*b^3-23*B*cos(d*x+c)^4*a^3-5*A*cos(d*x+c)^3*sin(d*x+c)*(cos(

d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c

),((a-b)/(a+b))^(1/2))*b^3-9*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+

cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+23*B*cos(d*x+c)^3*sin(d

*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c)

)/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+9*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(

d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-5*A*cos(d*x+

c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+

cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-9*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*

((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*E...

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sq

rt(b*sec(d*x + c) + a), x)

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

[Out]

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

[Out]

integrate((B*sec(d*x + c) + A)*(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 \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\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((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2),x)

[Out]

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

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