3.236 \(\int \frac {(a+a \sec (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx\)

Optimal. Leaf size=112 \[ \frac {5 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d} \]

[Out]

5*a^(5/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+a^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+
c))^(1/2)+a^2*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.22, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3814, 4015, 3801, 215} \[ \frac {a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{d}+\frac {5 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^(5/2)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])
/(d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d

Rule 215

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

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 4015

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx &=\frac {a^2 \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}+a \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a}{2}+\frac {5}{2} a \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{2} \left (5 a^2\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}-\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {5 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 91, normalized size = 0.81 \[ \frac {a^3 \tan (c+d x) \left ((2 \cos (c+d x)+1) \sqrt {(\cos (c+d x)-1) \sec ^2(c+d x)}+5 \sin ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a (\sec (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.39, size = 346, normalized size = 3.09 \[ \left [\frac {5 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {5 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(5*(a^2*cos(d*x + c) + a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*co
s(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x +
c)^3 + cos(d*x + c)^2)) + 4*(2*a^2*cos(d*x + c) + a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sq
rt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/2*(5*(a^2*cos(d*x + c) + a^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos
(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(2
*a^2*cos(d*x + c) + a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x +
c) + d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), x)

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maple [B]  time = 1.59, size = 199, normalized size = 1.78 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (5 \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}-5 \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+8 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )-4\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, a^{2}}{4 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x)

[Out]

-1/4/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(5*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2
^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*2^(1/2)-5*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+
cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+8*cos(d*x+c)^2-4*cos(d*x+c
)-4)*(1/cos(d*x+c))^(1/2)/sin(d*x+c)*a^2

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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((a+a*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(1/2),x)

[Out]

int((a + a/cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(1/2), x)

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sympy [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((a+a*sec(d*x+c))**(5/2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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