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

Optimal. Leaf size=89 \[ \frac {64 a^3 \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[Out]

2/5*a*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+64/15*a^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/15*a^2*(a+a*sec(d*x
+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3793, 3792} \[ \frac {64 a^3 \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d)
+ (2*a*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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]

Rule 3793

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

Rubi steps

\begin {align*} \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac {2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{5} (8 a) \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\\ &=\frac {16 a^2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{15} \left (32 a^2\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {64 a^3 \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 50, normalized size = 0.56 \[ \frac {2 a^3 \tan (c+d x) \left (3 \sec ^2(c+d x)+14 \sec (c+d x)+43\right )}{15 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(43 + 14*Sec[c + d*x] + 3*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.83, size = 82, normalized size = 0.92 \[ \frac {2 \, {\left (43 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(43*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 3*a^2)*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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giac [A]  time = 6.12, size = 122, normalized size = 1.37 \[ \frac {8 \, {\left (15 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, {\left (2 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 5 \, \sqrt {2} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/15*(15*sqrt(2)*a^5*sgn(cos(d*x + c)) + 4*(2*sqrt(2)*a^5*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)
*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a
*tan(1/2*d*x + 1/2*c)^2 + a)*d)

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maple [A]  time = 0.95, size = 75, normalized size = 0.84 \[ -\frac {2 \left (43 \left (\cos ^{3}\left (d x +c \right )\right )-29 \left (\cos ^{2}\left (d x +c \right )\right )-11 \cos \left (d x +c \right )-3\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a^{2}}{15 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/d*(43*cos(d*x+c)^3-29*cos(d*x+c)^2-11*cos(d*x+c)-3)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d
*x+c)^2*a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.51, size = 146, normalized size = 1.64 \[ -\frac {2\,a^2\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,15{}\mathrm {i}-{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,70{}\mathrm {i}+{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,70{}\mathrm {i}-{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,15{}\mathrm {i}+{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,43{}\mathrm {i}-43{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*15i - exp(c*2i + d*x
*2i)*70i + exp(c*3i + d*x*3i)*70i - exp(c*4i + d*x*4i)*15i + exp(c*5i + d*x*5i)*43i - 43i))/(15*d*(exp(c*1i +
d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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