\(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\) [538]

Optimal result

Integrand size = 23, antiderivative size = 342 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]

[Out]

4/105*a*(a-b)*(3*a^2-41*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^3/d+2/105*(a-b)*(6*a^2+57*a*b-25*b^2)
*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^2/d-4/35*a*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/7*(a+b*sec(d*x
+c))^(5/2)*tan(d*x+c)/b/d-2/105*(6*a^2-25*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3925, 4087, 4090, 3917, 4089} \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{105 b^2 d}+\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{105 b^3 d}-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{105 b d}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 b d} \]

[In]

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

[Out]

(4*a*(a - b)*Sqrt[a + b]*(3*a^2 - 41*b^2)*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^3*d) +
(2*(a - b)*Sqrt[a + b]*(6*a^2 + 57*a*b - 25*b^2)*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^
2*d) - (2*(6*a^2 - 25*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(105*b*d) - (4*a*(a + b*Sec[c + d*x])^(3/2)*
Tan[c + d*x])/(35*b*d) + (2*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*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 3925

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-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*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4087

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {2 \int \sec (c+d x) \left (\frac {5 b}{2}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{3/2} \, dx}{7 b} \\ & = -\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {4 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {19 a b}{4}-\frac {1}{4} \left (6 a^2-25 b^2\right ) \sec (c+d x)\right ) \, dx}{35 b} \\ & = -\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}+\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{8} b \left (51 a^2+25 b^2\right )-\frac {1}{4} a \left (3 a^2-41 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b} \\ & = -\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d}-\frac {\left (2 a \left (3 a^2-41 b^2\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b}+\frac {\left ((a-b) \left (6 a^2+57 a b-25 b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 b} \\ & = \frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) E\left (\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}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\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}}}{105 b^2 d}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 11.79 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 a \left (3 a^3+3 a^2 b-41 a b^2-41 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b \left (-6 a^3+51 a^2 b+82 a b^2+25 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (3 a^2-41 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^2 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {4 a \left (3 a^2-41 b^2\right ) \sin (c+d x)}{105 b^2}+\frac {2 \sec (c+d x) \left (3 a^2 \sin (c+d x)+25 b^2 \sin (c+d x)\right )}{105 b}+\frac {16}{35} a \sec (c+d x) \tan (c+d x)+\frac {2}{7} b \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \]

[In]

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

[Out]

(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*a*(3*a^3 + 3*a^2*b - 41*a*b^2 - 41*b^3)
*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[ArcSi
n[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-6*a^3 + 51*a^2*b + 82*a*b^2 + 25*b^3)*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)] + a*(3*a^2 - 41*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(105*b^2
*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3
/2)*((-4*a*(3*a^2 - 41*b^2)*Sin[c + d*x])/(105*b^2) + (2*Sec[c + d*x]*(3*a^2*Sin[c + d*x] + 25*b^2*Sin[c + d*x
]))/(105*b) + (16*a*Sec[c + d*x]*Tan[c + d*x])/35 + (2*b*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*
x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2342\) vs. \(2(308)=616\).

Time = 15.60 (sec) , antiderivative size = 2343, normalized size of antiderivative = 6.85

method	result	size
default	\(\text {Expression too large to display}\)	\(2343\)

[In]

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

[Out]

2/105/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-3*a^3*b*sin(d*x+c)+107*a*b^3*sin(d*x+c)+2
5*b^4*sin(d*x+c)+27*a^2*b^2*sin(d*x+c)-6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-
csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b+82*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+
c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2+82
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*a*b^3-25*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos(d*x+c)^2-6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(
d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*co
s(d*x+c)^2-50*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(
1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos(d*x+c)-12*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*cos(d*x+c)+6*(1/(a+b)*(b
+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*a^3*b-51*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/
(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2-82*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ell
ipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3+25*a*b^3*cos(d*x+c)*
sin(d*x+c)-6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/
2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4+39*a*b^3*tan(d*x+c)*sec(d*x+c)+3*a^3*b*cos(d*x+c)*sin(d*x+c)+82*a^2*
b^2*cos(d*x+c)*sin(d*x+c)-25*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(
(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4+25*tan(d*x+c)*b^4+82*(1/(a+b)*(b+a*cos(d*x+c))/(cos(
d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^
2*cos(d*x+c)^2+82*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(
cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*cos(d*x+c)^2+12*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))
^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x+c)-102
*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)-164*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3*cos(d*x+c)-12*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b)
)^(1/2))*a^3*b*cos(d*x+c)+164*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2*cos(d*x+c)+164*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d
*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*c
os(d*x+c)+6*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/
2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x+c)^2-51*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)^2-82*El
lipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*a*b^3*cos(d*x+c)^2-6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b*cos(d*x+c)^2+15*b^4*tan(d*x+c)*sec(
d*x+c)+15*b^4*tan(d*x+c)*sec(d*x+c)^2+27*a^2*b^2*tan(d*x+c)-6*a^4*cos(d*x+c)*sin(d*x+c)+39*a*b^3*tan(d*x+c))

Fricas [F]

\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

integral((b*sec(d*x + c)^4 + a*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**(3/2)*sec(c + d*x)**3, x)

Maxima [F]

\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

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

[Out]

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