3.2.36 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [136]

Optimal. Leaf size=103 \[ \frac {3 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}+\frac {3 a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{16 d}+\frac {\sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2754, 2746, 65, 212} \begin {gather*} \frac {3 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{4 d}+\frac {3 a \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 212

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

Rule 2746

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 2754

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

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Mathematica [A]
time = 0.19, size = 110, normalized size = 1.07 \begin {gather*} \frac {3 \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+2 a^2 (7-3 \sin (c+d x)) \sqrt {a (1+\sin (c+d x))}}{32 d (-1+\sin (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
^4 + 2*a^2*(7 - 3*Sin[c + d*x])*Sqrt[a*(1 + Sin[c + d*x])])/(32*d*(-1 + Sin[c + d*x])^2)

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Maple [A]
time = 0.62, size = 107, normalized size = 1.04

method	result	size
default	\(-\frac {2 a^{5} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{8 a \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {3 \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \left (a \sin \left (d x +c \right )-a \right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{8 a}\right )}{d}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^5*(-1/8*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a)^2-3/8/a*(-1/4*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a)
+1/8/a^(3/2)*2^(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 = 0.64, size = 134, normalized size = 1.30 \begin {gather*} -\frac {3 \, \sqrt {2} a^{\frac {7}{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 (3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} - 10 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}}{64 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(3*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x +
c) + a))) + 4*(3*(a*sin(d*x + c) + a)^(3/2)*a^4 - 10*sqrt(a*sin(d*x + c) + a)*a^5)/((a*sin(d*x + c) + a)^2 -
4*(a*sin(d*x + c) + a)*a + 4*a^2))/(a*d)

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Fricas [A]
time = 0.38, size = 147, normalized size = 1.43 \begin {gather*} \frac {3 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} a^{2} \sin \left (d x + c\right ) - 2 \, \sqrt {2} a^{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 (3 \, a^{2} \sin \left (d x + c\right ) - 7 \, a^{2}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(3*(sqrt(2)*a^2*cos(d*x + c)^2 + 2*sqrt(2)*a^2*sin(d*x + c) - 2*sqrt(2)*a^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*(3*a^2*sin(d*x + c) - 7*a^2)*sqrt
(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 5.09, size = 112, normalized size = 1.09 \begin {gather*} -\frac {\sqrt {2} a^{\frac {5}{2}} {\left (\frac {2 \, {\left (3 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - 3 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 3 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*sqrt(2)*a^(5/2)*(2*(3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 5*cos(-1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi
+ 1/2*d*x + 1/2*c)^2 - 1)^2 - 3*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 3*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c
) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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