3.125 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=127 \[ \frac {15 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}-\frac {15 a^2}{32 d \sqrt {a \sin (c+d x)+a}}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}+\frac {5 a \sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{16 d} \]

[Out]

1/4*sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2)/d+15/64*a^(3/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^
(1/2)/d-15/32*a^2/d/(a+a*sin(d*x+c))^(1/2)+5/16*a*sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.18, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2675, 2667, 51, 63, 206} \[ -\frac {15 a^2}{32 d \sqrt {a \sin (c+d x)+a}}+\frac {15 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}+\frac {5 a \sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(15*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(32*Sqrt[2]*d) - (15*a^2)/(32*d*Sqrt[a + a*Si
n[c + d*x]]) + (5*a*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/
2))/(4*d)

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {1}{8} (5 a) \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {1}{32} \left (15 a^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {\left (15 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=-\frac {15 a^2}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {\left (15 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac {15 a^2}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}+\frac {\left (15 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{32 d}\\ &=\frac {15 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}-\frac {15 a^2}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {5 a \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{4 d}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 44, normalized size = 0.35 \[ -\frac {a^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{4 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/4*(a^2*Hypergeometric2F1[-1/2, 3, 1/2, (1 + Sin[c + d*x])/2])/(d*Sqrt[a + a*Sin[c + d*x]])

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fricas [A]  time = 0.77, size = 155, normalized size = 1.22 \[ \frac {15 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - \sqrt {2} a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, a \cos \left (d x + c\right )^{2} + 20 \, a \sin \left (d x + c\right ) - 12 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{128 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/128*(15*(sqrt(2)*a*cos(d*x + c)^2*sin(d*x + c) - sqrt(2)*a*cos(d*x + c)^2)*sqrt(a)*log(-(a*sin(d*x + c) + 2*
sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) - 4*(15*a*cos(d*x + c)^2 + 20*a*sin(d*x +
c) - 12*a)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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giac [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(sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.33, size = 101, normalized size = 0.80 \[ -\frac {2 a^{5} \left (\frac {1}{8 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (7 \sin \left (d x +c \right )-11\right )}{8 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}}{8 a^{3}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2*a^5*(1/8/a^3/(a+a*sin(d*x+c))^(1/2)+1/8/a^3*(1/8*(a+a*sin(d*x+c))^(1/2)*a*(7*sin(d*x+c)-11)/(a*sin(d*x+c)-a
)^2-15/16*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d

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maxima [A]  time = 0.61, size = 151, normalized size = 1.19 \[ -\frac {15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} - 50 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 32 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}}}{128 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/128*(15*sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x
+ c) + a))) + 4*(15*(a*sin(d*x + c) + a)^2*a^3 - 50*(a*sin(d*x + c) + a)*a^4 + 32*a^5)/((a*sin(d*x + c) + a)^
(5/2) - 4*(a*sin(d*x + c) + a)^(3/2)*a + 4*sqrt(a*sin(d*x + c) + a)*a^2))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^5,x)

[Out]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^5, 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(sec(d*x+c)**5*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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