3.162 \(\int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 (5 c+8 d) \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{15 f}+\frac{2 a^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}+\frac{2 a d \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{5 f} \]

[Out]

(2*a^(5/2)*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f + (2*a^3*(35*c + 32*d)*Tan[e + f*x])/(
15*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(5*c + 8*d)*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(15*f) + (2*a*d*(a
+ a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f)

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Rubi [A]  time = 0.231934, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3917, 3915, 3774, 203, 3792} \[ \frac{2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 (5 c+8 d) \tan (e+f x) \sqrt{a \sec (e+f x)+a}}{15 f}+\frac{2 a^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}+\frac{2 a d \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x]),x]

[Out]

(2*a^(5/2)*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f + (2*a^3*(35*c + 32*d)*Tan[e + f*x])/(
15*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(5*c + 8*d)*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(15*f) + (2*a*d*(a
+ a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f)

Rule 3917

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] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3915

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 3774

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 203

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

Rule 3792

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]

Rubi steps

\begin{align*} \int (a+a \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx &=\frac{2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac{2}{5} \int (a+a \sec (e+f x))^{3/2} \left (\frac{5 a c}{2}+\frac{1}{2} a (5 c+8 d) \sec (e+f x)\right ) \, dx\\ &=\frac{2 a^2 (5 c+8 d) \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac{2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac{4}{15} \int \sqrt{a+a \sec (e+f x)} \left (\frac{15 a^2 c}{4}+\frac{1}{4} a^2 (35 c+32 d) \sec (e+f x)\right ) \, dx\\ &=\frac{2 a^2 (5 c+8 d) \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac{2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\left (a^2 c\right ) \int \sqrt{a+a \sec (e+f x)} \, dx+\frac{1}{15} \left (a^2 (35 c+32 d)\right ) \int \sec (e+f x) \sqrt{a+a \sec (e+f x)} \, dx\\ &=\frac{2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (5 c+8 d) \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac{2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}-\frac{\left (2 a^3 c\right ) \operatorname{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^{5/2} c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}+\frac{2 a^3 (35 c+32 d) \tan (e+f x)}{15 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (5 c+8 d) \sqrt{a+a \sec (e+f x)} \tan (e+f x)}{15 f}+\frac{2 a d (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.921576, size = 128, normalized size = 0.9 \[ \frac{a^2 \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (2 \sin \left (\frac{1}{2} (e+f x)\right ) (2 (5 c+14 d) \cos (e+f x)+(40 c+43 d) \cos (2 (e+f x))+40 c+49 d)+30 \sqrt{2} c \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \cos ^{\frac{5}{2}}(e+f x)\right )}{30 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x]),x]

[Out]

(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*(30*Sqrt[2]*c*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]
*Cos[e + f*x]^(5/2) + 2*(40*c + 49*d + 2*(5*c + 14*d)*Cos[e + f*x] + (40*c + 43*d)*Cos[2*(e + f*x)])*Sin[(e +
f*x)/2]))/(30*f)

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Maple [B]  time = 0.241, size = 341, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2}}{60\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 15\,\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}c+30\,\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ){\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}c+15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \sqrt{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}c\sin \left ( fx+e \right ) +320\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}c+344\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}d-280\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}c-232\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}d-40\,c\cos \left ( fx+e \right ) -88\,d\cos \left ( fx+e \right ) -24\,d \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x)

[Out]

-1/60/f*a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(15*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*arctanh(1/2*2^(1/2)*(-2*
cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*c+30*2^(1/2)*sin(
f*x+e)*cos(f*x+e)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+
e)/(1+cos(f*x+e)))^(5/2)*c+15*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*
2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*c*sin(f*x+e)+320*cos(f*x+e)^3*c+344*cos(f*x+e)^3*d-280*cos(f*x+e)
^2*c-232*cos(f*x+e)^2*d-40*c*cos(f*x+e)-88*d*cos(f*x+e)-24*d)/sin(f*x+e)/cos(f*x+e)^2

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Maxima [B]  time = 1.97029, size = 1885, normalized size = 13.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="maxima")

[Out]

1/6*(30*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/
4)*((12*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(2*f*x + 2*e) - 3*a^2*sin(2*f*x + 2*e) - 4
*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(3/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (12*a^2*sin(2*f*x + 2*e)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) + 3*a^2*cos(2*f*x + 2*e) - a^2 + 4*(3*a^2*cos(2*f*x + 2*e) + 4*a^2)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sqrt(a) + 3*((a^2*cos(2*f*x + 2*e)^
2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*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*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)*(c
os(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^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*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*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^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2*e)^2 + 2*a^2*cos(2*f*x + 2
*e) + a^2)*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(si
n(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^2*cos(2*f*x + 2*e)^2 + a^2*sin(2*f*x + 2
*e)^2 + 2*a^2*cos(2*f*x + 2*e) + a^2)*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))*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)*f)

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Fricas [A]  time = 1.0254, size = 976, normalized size = 6.87 \begin{align*} \left [\frac{15 \,{\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\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 (3 \, a^{2} d +{\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} +{\left (5 \, a^{2} c + 14 \, a^{2} 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 )}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac{2 \,{\left (15 \,{\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )^{2}\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 (3 \, a^{2} d +{\left (40 \, a^{2} c + 43 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} +{\left (5 \, a^{2} c + 14 \, a^{2} 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 )}}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[1/15*(15*(a^2*c*cos(f*x + e)^3 + a^2*c*cos(f*x + e)^2)*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*(3*a^2
*d + (40*a^2*c + 43*a^2*d)*cos(f*x + e)^2 + (5*a^2*c + 14*a^2*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)^3 + f*cos(f*x + e)^2), -2/15*(15*(a^2*c*cos(f*x + e)^3 + a^2*c*cos(f*x
+ e)^2)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (3*a^2*d
+ (40*a^2*c + 43*a^2*d)*cos(f*x + e)^2 + (5*a^2*c + 14*a^2*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)^3 + f*cos(f*x + e)^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)*(c+d*sec(f*x+e)),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="giac")

[Out]

Timed out