3.2.12 \(\int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [112]

Optimal. Leaf size=149 \[ \frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d} \]

[Out]

-35/96*a^2/d/(a+a*sin(d*x+c))^(3/2)+35/128*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)
/d-35/64*a/d/(a+a*sin(d*x+c))^(1/2)+7/16*a*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)+1/4*sec(d*x+c)^4*(a+a*sin(d*x
+c))^(1/2)/d

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Rubi [A]
time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2754, 2766, 2746, 53, 65, 212} \begin {gather*} -\frac {35 a^2}{96 d (a \sin (c+d x)+a)^{3/2}}-\frac {35 a}{64 d \sqrt {a \sin (c+d x)+a}}+\frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a \sin (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(35*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(64*Sqrt[2]*d) - (35*a^2)/(96*d*(a + a*Sin[c
+ d*x])^(3/2)) - (35*a)/(64*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*Sec[c + d*x]^2)/(16*d*Sqrt[a + a*Sin[c + d*x]])
+ (Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(4*d)

Rule 53

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 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]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{8} (7 a) \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{32} \left (35 a^2\right ) \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (35 a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (35 a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{64 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {(35 a) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {(35 a) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{64 d}\\ &=\frac {35 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.32, size = 179, normalized size = 1.20 \begin {gather*} \frac {\sqrt {a (1+\sin (c+d x))} \left ((-420+420 i) \sqrt [4]{-1} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )+\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+\frac {-102-70 \cos (2 (c+d x))+329 \sin (c+d x)+105 \sin (3 (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}\right )}{768 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((-420 + 420*I)*(-1)^(1/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c +
d*x)/4] + Sin[(2*c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (-102 - 70*Cos[2*(c + d*x)] + 329*Si
n[c + d*x] + 105*Sin[3*(c + d*x)])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4))/(768*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^4)

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Maple [A]
time = 0.79, size = 118, normalized size = 0.79

method	result	size
default	\(-\frac {2 a^{5} \left (\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (11 \sin \left (d x +c \right )-15\right )}{8 \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {35 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}}{16 a^{4}}+\frac {3}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^5*(1/16/a^4*(1/8*(a+a*sin(d*x+c))^(1/2)*a*(11*sin(d*x+c)-15)/(a*sin(d*x+c)-a)^2-35/16*2^(1/2)/a^(1/2)*arc
tanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))+3/16/a^4/(a+a*sin(d*x+c))^(1/2)+1/24/a^3/(a+a*sin(d*x+c))^(3
/2))/d

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Maxima [A]
time = 0.52, size = 168, normalized size = 1.13 \begin {gather*} -\frac {105 \, \sqrt {2} a^{\frac {3}{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 (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} - 350 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 224 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 64 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}}}{768 \, 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))^(1/2),x, algorithm="maxima")

[Out]

-1/768*(105*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*
x + c) + a))) + 4*(105*(a*sin(d*x + c) + a)^3*a^2 - 350*(a*sin(d*x + c) + a)^2*a^3 + 224*(a*sin(d*x + c) + a)*
a^4 + 64*a^5)/((a*sin(d*x + c) + a)^(7/2) - 4*(a*sin(d*x + c) + a)^(5/2)*a + 4*(a*sin(d*x + c) + a)^(3/2)*a^2)
)/(a*d)

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Fricas [A]
time = 0.37, size = 121, normalized size = 0.81 \begin {gather*} \frac {105 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{4} \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 (35 \, \cos \left (d x + c\right )^{2} - 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, d \cos \left (d x + c\right )^{4}} \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))^(1/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(2)*sqrt(a)*cos(d*x + c)^4*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*(35*cos(d*x + c)^2 - 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x +
c) + a))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sec ^{5}{\left (c + d x \right )}\, dx \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))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**5, x)

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Giac [A]
time = 6.90, size = 146, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {2} \sqrt {a} {\left (\frac {6 \, {\left (11 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \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}} + \frac {16 \, {\left (9 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - 105 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 105 \, \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 )}{768 \, 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))^(1/2),x, algorithm="giac")

[Out]

-1/768*sqrt(2)*sqrt(a)*(6*(11*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 13*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 + 16*(9*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 1)/cos(-1/4*pi + 1/2*d*x + 1/2*c)^3
- 105*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 105*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 {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\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))^(1/2)/cos(c + d*x)^5,x)

[Out]

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

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