\(\int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx\) [156]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\frac {2 a^{3/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^2 (3 c+4 d) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a d \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{3 f} \]

[Out]

2*a^(3/2)*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/f+2/3*a^2*(3*c+4*d)*tan(f*x+e)/f/(a+a*sec(f*x+e)
)^(1/2)+2/3*a*d*(a+a*sec(f*x+e))^(1/2)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4002, 4000, 3859, 209, 3877} \[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\frac {2 a^{3/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}+\frac {2 a^2 (3 c+4 d) \tan (e+f x)}{3 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a d \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{3 f} \]

[In]

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

[Out]

(2*a^(3/2)*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f + (2*a^2*(3*c + 4*d)*Tan[e + f*x])/(3*
f*Sqrt[a + a*Sec[e + f*x]]) + (2*a*d*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(3*f)

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a d \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{3 f}+\frac {2}{3} \int \sqrt {a+a \sec (e+f x)} \left (\frac {3 a c}{2}+\frac {1}{2} a (3 c+4 d) \sec (e+f x)\right ) \, dx \\ & = \frac {2 a d \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{3 f}+(a c) \int \sqrt {a+a \sec (e+f x)} \, dx+\frac {1}{3} (a (3 c+4 d)) \int \sec (e+f x) \sqrt {a+a \sec (e+f x)} \, dx \\ & = \frac {2 a^2 (3 c+4 d) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a d \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{3 f}-\frac {\left (2 a^2 c\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f} \\ & = \frac {2 a^{3/2} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {2 a^2 (3 c+4 d) \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a d \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.97 \[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\frac {a \sec \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {a (1+\sec (e+f x))} \left (3 \sqrt {2} c \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {3}{2}}(e+f x)+2 (d+(3 c+5 d) \cos (e+f x)) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{3 f} \]

[In]

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

[Out]

(a*Sec[(e + f*x)/2]*Sec[e + f*x]*Sqrt[a*(1 + Sec[e + f*x])]*(3*Sqrt[2]*c*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]*Cos[
e + f*x]^(3/2) + 2*(d + (3*c + 5*d)*Cos[e + f*x])*Sin[(e + f*x)/2]))/(3*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(91)=182\).

Time = 1.46 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.96

method result size
default \(\frac {2 c a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\sin \left (f x +e \right )\right )}{f \left (\cos \left (f x +e \right )+1\right )}+\frac {2 d a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (5 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}\) \(206\)
parts \(\frac {2 c a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\sin \left (f x +e \right )\right )}{f \left (\cos \left (f x +e \right )+1\right )}+\frac {2 d a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (5 \sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{3 f \left (\cos \left (f x +e \right )+1\right )}\) \(206\)

[In]

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

[Out]

2*c/f*a*(a*(sec(f*x+e)+1))^(1/2)*(arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(
f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*
(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+sin(f*x+e))/(cos(f*x+e)+1)+2/3*d/f*a*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1
)*(5*sin(f*x+e)+tan(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.01 \[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\left [\frac {3 \, {\left (a c \cos \left (f x + e\right )^{2} + a c \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \, {\left (a d + {\left (3 \, a c + 5 \, a d\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (a c \cos \left (f x + e\right )^{2} + a c \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - {\left (a d + {\left (3 \, a c + 5 \, a d\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{3 \, {\left (f \cos \left (f x + e\right )^{2} + f \cos \left (f x + e\right )\right )}}\right ] \]

[In]

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

[Out]

[1/3*(3*(a*c*cos(f*x + e)^2 + a*c*cos(f*x + e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 2*(a*d + (3*a*c
+ 5*a*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^2 + f*cos(f*x + e
)), -2/3*(3*(a*c*cos(f*x + e)^2 + a*c*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos
(f*x + e)/(sqrt(a)*sin(f*x + e))) - (a*d + (3*a*c + 5*a*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e
))*sin(f*x + e))/(f*cos(f*x + e)^2 + f*cos(f*x + e))]

Sympy [F]

\[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sec {\left (e + f x \right )}\right )\, dx \]

[In]

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

[Out]

Integral((a*(sec(e + f*x) + 1))**(3/2)*(c + d*sec(e + f*x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 998 vs. \(2 (91) = 182\).

Time = 0.39 (sec) , antiderivative size = 998, normalized size of antiderivative = 9.50 \[ \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*((a*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e
)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) +
 1))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
 + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))) + 1) - a*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x
+ 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) +
1)^(1/4)*(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
 + 2*e))) + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e)))) - 1) - a*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*
e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) + a*arctan2((cos(2*f*x + 2*e)^2 +
sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (
cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e) + 1)) - 1))*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sqrt(a) + 4*
(a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)) - (a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - a)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e) + 1)))*sqrt(a))*c/((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*f)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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