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

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

Optimal. Leaf size=384 \[ \frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 (a-b) \sqrt {a+b} \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b d}-\frac {2 (a-b) \sqrt {a+b} \left (15 a^3 B+161 a^2 A b+145 a b^2 B+63 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b^2 d}+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]

[Out]

-2/105*(a-b)*(161*A*a^2*b+63*A*b^3+15*B*a^3+145*B*a*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^2/d+2/105

*(a-b)*(b^2*(63*A-25*B)-8*a*b*(7*A-15*B)+15*a^2*(7*A-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+2/35*(7

*A*b+5*B*a)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*B*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/105*(56*A*a*b+15*B

*a^2+25*B*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A] time = 0.81, antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4002, 4005, 3832, 4004} \[ \frac {2 \left (15 a^2 B+56 a A b+25 b^2 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 (a-b) \sqrt {a+b} \left (15 a^2 (7 A-B)-8 a b (7 A-15 B)+b^2 (63 A-25 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b d}-\frac {2 (a-b) \sqrt {a+b} \left (161 a^2 A b+15 a^3 B+145 a b^2 B+63 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b^2 d}+\frac {2 (5 a B+7 A b) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*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))])/(105*b^2*d) + (2*(a - b)*Sqrt[a + b]*(b^2*(63*A - 25*B) - 8*a*b*(7*A - 15*B) + 15*a^2*(7*A -

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))])/(105*b*d) + (2*(56*a*A*b + 15*a^2*B + 25*b^2*B)*Sqr

t[a + b*Sec[c + d*x]]*Tan[c + d*x])/(105*d) + (2*(7*A*b + 5*a*B)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*

d) + (2*B*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d)

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr

t[(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)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))

, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x

]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr

eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4004

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 + Cs

c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]

)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4005

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]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {2}{7} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (7 a A+5 b B)+\frac {1}{2} (7 A b+5 a B) \sec (c+d x)\right ) \, dx\\ &=\frac {2 (7 A b+5 a B) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {4}{35} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {1}{4} \left (35 a^2 A+21 A b^2+40 a b B\right )+\frac {1}{4} \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {8}{105} \int \frac {\sec (c+d x) \left (\frac {1}{8} \left (105 a^3 A+119 a A b^2+135 a^2 b B+25 b^3 B\right )+\frac {1}{8} \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}+\frac {1}{105} \left ((a-b) \left (b^2 (63 A-25 B)-8 a b (7 A-15 B)+15 a^2 (7 A-B)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{105} \left (161 a^2 A b+63 A b^3+15 a^3 B+145 a b^2 B\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 (161 a^2 A b+63 A b^3+15 a^3 B+145 a 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}}}{105 b^2 d}+\frac {2 (a-b) \sqrt {a+b} \left (b^2 (63 A-25 B)-8 a b (7 A-15 B)+15 a^2 (7 A-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}}}{105 b d}+\frac {2 \left (56 a A b+15 a^2 B+25 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A b+5 a B) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [B] time = 23.17, size = 2957, normalized size = 7.70 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x])*((2*(161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a

*b^2*B)*Sin[c + d*x])/(105*b) + (2*Sec[c + d*x]^2*(7*A*b^2*Sin[c + d*x] + 15*a*b*B*Sin[c + d*x]))/35 + (2*Sec[

c + d*x]*(77*a*A*b*Sin[c + d*x] + 45*a^2*B*Sin[c + d*x] + 25*b^2*B*Sin[c + d*x]))/105 + (2*b^2*B*Sec[c + d*x]^

2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x])^2*(B + A*Cos[c + d*x])) + (2*((-23*a^2*A*b)/(15*Sqrt[b + a*Cos[c +

d*x]]*Sqrt[Sec[c + d*x]]) - (3*A*b^3)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^3*B)/(7*Sqrt[b + a

*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (29*a*b^2*B)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^3*A*

Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (8*a*A*b^2*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]

]) - (a^4*B*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (2*a^2*b*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a

*Cos[c + d*x]]) + (5*b^3*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (23*a^3*A*Cos[2*(c + d*x)]*Sqrt

[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) - (3*a*A*b^2*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*Sqrt[b + a*

Cos[c + d*x]]) - (a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (29*a^2*b*B*Cos[

2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*S

ec[c + d*x])^(5/2)*(A + B*Sec[c + d*x])*((-2*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*((161*a^2*A*b + 63*A*b^3

+ 15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*(7*A + B) + 8*a*b*(

7*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x])/Sqrt[(b +

a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Tan[(c + d*

x)/2]*(-1 + Tan[(c + d*x)/2]^2)))/(105*b*d*(b + a*Cos[c + d*x])^2*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2

]*Sec[c + d*x]^(7/2)*(-1/105*(a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*((-2*(Cos[c + d*x]/(1 + Cos

[c + d*x]))^(3/2)*((161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a -

b)/(a + b)] - b*(15*a^2*(7*A + B) + 8*a*b*(7*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]]

, (a - b)/(a + b)])*Sec[c + d*x])/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (161*a^2*A*b + 63*

A*b^3 + 15*a^3*B + 145*a*b^2*B)*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)))/(b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[

Sec[(c + d*x)/2]^2]) - (Sqrt[b + a*Cos[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*((-2*(

Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*((161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan

[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*(7*A + B) + 8*a*b*(7*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[Arc

Sin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x])/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]

+ (161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2)))/(105*b*Sqrt[

Sec[(c + d*x)/2]^2]) + (Sqrt[b + a*Cos[c + d*x]]*((-2*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*((161*a^2*A*b +

63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*(7*A + B)

+ 8*a*b*(7*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x])/

Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Ta

n[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]^2))*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2

]^2*Sec[c + d*x]*Tan[c + d*x]))/(105*b*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]) + (2*Sq

rt[b + a*Cos[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*((-3*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*((161*

a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*

(7*A + B) + 8*a*b*(7*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[

c + d*x]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[(b + a*Cos

[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + ((Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*((161*a^2*A*b + 63*A*b^3 +

15*a^3*B + 145*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*(7*A + B) + 8*a*b*(7

*A + 15*B) + b^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x]*(-((a*Sin[c

+ d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/

((b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x])))^(3/2) + (161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*

Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + ((161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sec[(c + d*x)/2]^2*

(-1 + Tan[(c + d*x)/2]^2))/2 - (2*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*Sec[c + d*x]*(-1/2*(b*(15*a^2*(7*A +

B) + 8*a*b*(7*A + 15*B) + b^2*(63*A + 25*B))*Sec[(c + d*x)/2]^2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a -

b)*Tan[(c + d*x)/2]^2)/(a + b)]) + ((161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 145*a*b^2*B)*Sec[(c + d*x)/2]^2*Sqrt

[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])/(2*Sqrt[1 - Tan[(c + d*x)/2]^2])))/Sqrt[(b + a*Cos[c + d*x])/((a +

b)*(1 + Cos[c + d*x]))] - (2*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*((161*a^2*A*b + 63*A*b^3 + 15*a^3*B + 14

5*a*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - b*(15*a^2*(7*A + B) + 8*a*b*(7*A + 15*B) + b

^2*(63*A + 25*B))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[c + d*x]*Tan[c + d*x])/Sqrt[(b + a

*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]))/(105*b*Sqrt[Sec[(c + d*x)/2]^2])))

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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{2} \sec \left (d x + c\right )^{4} + A a^{2} \sec \left (d x + c\right ) + {\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{3} + {\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)^4 + A*a^2*sec(d*x + c) + (2*B*a*b + A*b^2)*sec(d*x + c)^3 + (B*a^2 + 2*A*a*b)*sec

(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)*sec(d*x + c), x)

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maple [B] time = 2.69, size = 3637, normalized size = 9.47 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105/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*(-161*A*cos(d*x+c)^5*a^3*b-77*A

*cos(d*x+c)^5*a^2*b^2-63*A*cos(d*x+c)^5*a*b^3-45*B*cos(d*x+c)^5*a^3*b-145*B*cos(d*x+c)^5*a^2*b^2+161*A*cos(d*x

+c)^4*a^3*b-161*A*cos(d*x+c)^4*a^2*b^2-15*B*cos(d*x+c)^4*a^3*b+238*A*cos(d*x+c)^3*a^2*b^2+60*B*cos(d*x+c)^3*a^

3*b+110*B*cos(d*x+c)^3*a*b^3+90*B*cos(d*x+c)^2*a^2*b^2+60*B*cos(d*x+c)*a*b^3+98*A*cos(d*x+c)^2*a*b^3+42*A*cos(

d*x+c)^3*b^4-25*B*cos(d*x+c)^4*b^4+10*B*cos(d*x+c)^2*b^4-145*B*sin(d*x+c)*cos(d*x+c)^4*(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^3-35*A*cos(d*x+c)^4*a*b^3-25*B*cos(d*x+c)^5*a*b^3+55*B*cos(d*x+c)^4*a^2*b^2+15*B*b^4+15*B*sin(d*x+c)*

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)*EllipticE((-1+cos

(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+145*B*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b

^2+145*B*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-15*B*sin(d*x+c)*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))*a^3*b-135*B*sin(d*x+c)*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))*a^2*b^2-145*B*sin(d*x+c)*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))*a*b^3+161*A*sin(d*x+c)*cos(d*x+c)^4*(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*b+161*A*

sin(d*x+c)*cos(d*x+c)^4*(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)*Ellipt

icE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+63*A*sin(d*x+c)*cos(d*x+c)^4*(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^3-161*A*sin(d*x+c)*cos(d*x+c)^4*(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^2-119*A*sin(d*x+c)*cos(d*x+c)^4*(

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^3+15*B*sin(d*x+c)*cos(d*x+c)^4*(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*b+145*B*sin(d*x+

c)*cos(d*x+c)^4*(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^2+145*B*sin(d*x+c)*cos(d*x+c)^4*(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^3-15*B*sin(d*x+c)*cos(d*x+c)^4*(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^3*b-135*B*sin(d*x+c)*cos(d*x+c)^4*(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^2-15*B*cos(d*x+c)^5*a^4-105*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*b

-105*A*cos(d*x+c)^4*(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*b+161*A*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(

a+b))^(1/2))*a^3*b+161*A*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+63*A*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/

sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-161*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*co

s(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^2-119*A*

sin(d*x+c)*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)*Ellipt

icF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-63*A*cos(d*x+c)^4*b^4+15*B*cos(d*x+c)^4*a^4+21*A*cos

(d*x+c)*b^4+63*A*sin(d*x+c)*cos(d*x+c)^4*(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^4-63*A*sin(d*x+c)*cos(d*x+c)^4*(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^4+15*B*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+co

s(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4-25*B*sin(d*x+c)*cos(d*x+c

)^4*(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^4+63*A*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-63*A*sin(d*x+

c)*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))*b^4+15*B*sin(d*x+c)*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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4-2

5*B*sin(d*x+c)*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)*El

lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-145*B*cos(d*x+c)^4*a*b^3)/(b+a*cos(d*x+c))/cos(d*x

+c)^3/sin(d*x+c)^5/b

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

[Out]

Timed out

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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]

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))/cos(c + d*x),x)

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)*sec(c + d*x), x)

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