3.5.94 \(\int (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [494]

Optimal. Leaf size=142 \[ \frac {2 a^{3/2} A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

2*a^(3/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/5*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/15
*a^2*(15*A+20*B+12*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a*(5*B+3*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]
time = 0.17, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4139, 4002, 4000, 3859, 209, 3877} \begin {gather*} \frac {2 a^{3/2} A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 B+3 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(3/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(15*A + 20*B + 12*C)*Tan[c +
d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(5*B + 3*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*C*
(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 209

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

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3877

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

Rule 4000

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*Csc[e + f*x], x], x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4002

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

Rule 4139

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/(b*(m + 1)
), Int[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b
, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int (a+a \sec (c+d x))^{3/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (5 B+3 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4 \int \sqrt {a+a \sec (c+d x)} \left (\frac {15 a^2 A}{4}+\frac {1}{4} a^2 (15 A+20 B+12 C) \sec (c+d x)\right ) \, dx}{15 a}\\ &=\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+(a A) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{15} (a (15 A+20 B+12 C)) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac {\left (2 a^2 A\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 a^{3/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 (15 A+20 B+12 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (5 B+3 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 1.47, size = 132, normalized size = 0.93 \begin {gather*} \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (30 \sqrt {2} A \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (15 A+25 B+24 C+2 (5 B+9 C) \cos (c+d x)+(15 A+25 B+18 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(30*Sqrt[2]*A*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*C
os[c + d*x]^(5/2) + 2*(15*A + 25*B + 24*C + 2*(5*B + 9*C)*Cos[c + d*x] + (15*A + 25*B + 18*C)*Cos[2*(c + d*x)]
)*Sin[(c + d*x)/2]))/(30*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(360\) vs. \(2(124)=248\).
time = 16.29, size = 361, normalized size = 2.54

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (15 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+30 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+15 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+120 A \left (\cos ^{3}\left (d x +c \right )\right )+200 B \left (\cos ^{3}\left (d x +c \right )\right )+144 C \left (\cos ^{3}\left (d x +c \right )\right )-120 A \left (\cos ^{2}\left (d x +c \right )\right )-160 B \left (\cos ^{2}\left (d x +c \right )\right )-72 C \left (\cos ^{2}\left (d x +c \right )\right )-40 B \cos \left (d x +c \right )-48 C \cos \left (d x +c \right )-24 C \right ) a}{60 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}\)	\(361\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(15*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+30*A*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c
)*sin(d*x+c)*2^(1/2)+15*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+120*A*cos(d*x+c)^3+200*B*cos(d*x+c)^3+144*C*cos(d*x+c)^3-1
20*A*cos(d*x+c)^2-160*B*cos(d*x+c)^2-72*C*cos(d*x+c)^2-40*B*cos(d*x+c)-48*C*cos(d*x+c)-24*C)/cos(d*x+c)^2/sin(
d*x+c)*a

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

[Out]

-1/30*(15*((A*a*cos(2*d*x + 2*c)^2 + A*a*sin(2*d*x + 2*c)^2 + 2*A*a*cos(2*d*x + 2*c) + A*a)*arctan2((cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (A*a*cos(2*d*x + 2*c)^2 + A*a*sin(2*d*x + 2*c)^2 + 2*A*a*cos(2*d*x + 2*c
) + A*a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)
*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - 2*(A*a*d*cos(2*d*x + 2*c)^2 + A*a*d*sin(2*d*x
+ 2*c)^2 + 2*A*a*d*cos(2*d*x + 2*c) + A*a*d)*integrate((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*
c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*
x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(7/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(
6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*si
n(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*
cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c
)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(3/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/((cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 + (cos(2*d*x + 2*c)^2 + s
in(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c)^2 + 2*cos(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*sin(4*d*x + 4*c)^2 + (2*cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x +
2*c)^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 4*(cos(2*d*x + 2*c)^3 + c
os(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(4*d*x + 4*c) + cos(2*d*x + 2
*c)^2 + 2*(sin(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x
+ 4*c) + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 4*(sin(2*d*x + 2*
c)^3 + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(4*d*x + 4*c))*cos(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 + (cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co
s(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c)^2 + 2*cos(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c
os(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c)^2 + 4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
*sin(4*d*x + 4*c)^2 + (2*cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x + 2*c)
^3 + cos(2*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)*cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 4*(cos(2*d*x + 2*c)^3 + cos(2
*d*x + 2*c)*sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)^2 + cos(2*d*x + 2*c))*cos(4*d*x + 4*c) + cos(2*d*x + 2*c)^
2 + 2*(sin(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4
*c) + (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 4*(sin(2*d*x + 2*c)^3
+ (cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c))*sin(4*d*x + 4*c))*sin(3/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1))^2), x) - 2*((5*A + 6*B + 4*C)*a*d*cos(2*d*x + 2*c)^2 + (5*A + 6*B + 4*C)*a*d*s
in(2*d*x + 2*c)^2 + 2*(5*A + 6*B + 4*C)*a*d*cos(2*d*x + 2*c) + (5*A + 6*B + 4*C)*a*d)*integrate((cos(2*d*x + 2
*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x
+ 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*
x + 2*c) + sin(2*d*x + 2*c)^2)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*
d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(
2*d*x + 2*c))*sin(5/2*arctan2(sin(2*d*x + 2*c),...

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Fricas [A]
time = 2.45, size = 364, normalized size = 2.56 \begin {gather*} \left [\frac {15 \, {\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (15 \, A + 25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (A a \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (15 \, A + 25 \, B + 18 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (5 \, B + 9 \, C\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(A*a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((15*A + 2
5*B + 18*C)*a*cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin
(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), -2/15*(15*(A*a*cos(d*x + c)^3 + A*a*cos(d*x + c)^2)*sqrt(a)*
arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((15*A + 25*B + 18*C)*a*
cos(d*x + c)^2 + (5*B + 9*C)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)^3 + d*cos(d*x + c)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**(3/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (124) = 248\).
time = 1.51, size = 342, normalized size = 2.41 \begin {gather*} -\frac {\frac {15 \, A \sqrt {-a} a^{2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left ({\left (\sqrt {2} {\left (15 \, A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 20 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 12 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \sqrt {2} {\left (3 \, A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*A*sqrt(-a)*a^2*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4
*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(
2)*abs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) - 2*((sqrt(2)*(15*A*a^4*sgn(cos(d*x + c)) + 20*B*a^4*sgn(cos(d*x +
c)) + 12*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 10*sqrt(2)*(3*A*a^4*sgn(cos(d*x + c)) + 5*B*a^4*sgn
(cos(d*x + c)) + 3*C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(A*a^4*sgn(cos(d*x + c)) + 2*
B*a^4*sgn(cos(d*x + c)) + 2*C*a^4*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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