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

Optimal. Leaf size=150 \[ \frac {7 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {7}{16 a d \sqrt {a \sin (c+d x)+a}}-\frac {7}{24 d (a \sin (c+d x)+a)^{3/2}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a \sin (c+d x)+a}}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

-7/24/d/(a+a*sin(d*x+c))^(3/2)-1/5*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)+7/32*arctanh(1/2*(a+a*sin(d*x+c))^(1/
2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)-7/16/a/d/(a+a*sin(d*x+c))^(1/2)+7/20*sec(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(
1/2)

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Rubi [A]  time = 0.20, antiderivative size = 150, 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 = {2681, 2687, 2667, 51, 63, 206} \[ \frac {7 \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {7}{16 a d \sqrt {a \sin (c+d x)+a}}-\frac {7}{24 d (a \sin (c+d x)+a)^{3/2}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a \sin (c+d x)+a}}-\frac {\sec ^2(c+d x)}{5 d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(16*Sqrt[2]*a^(3/2)*d) - 7/(24*d*(a + a*Sin[c + d*x])^
(3/2)) - Sec[c + d*x]^2/(5*d*(a + a*Sin[c + d*x])^(3/2)) - 7/(16*a*d*Sqrt[a + a*Sin[c + d*x]]) + (7*Sec[c + d*
x]^2)/(20*a*d*Sqrt[a + a*Sin[c + d*x]])

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 2681

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*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2687

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)), Int[
(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 \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}+\frac {7 \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{10 a}\\ &=-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}+\frac {7}{8} \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}+\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=-\frac {7}{24 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=-\frac {7}{24 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}-\frac {7}{16 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{32 a d}\\ &=-\frac {7}{24 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}-\frac {7}{16 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{16 a d}\\ &=\frac {7 \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {7}{24 d (a+a \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x)}{5 d (a+a \sin (c+d x))^{3/2}}-\frac {7}{16 a d \sqrt {a+a \sin (c+d x)}}+\frac {7 \sec ^2(c+d x)}{20 a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 42, normalized size = 0.28 \[ -\frac {a \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{10 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*(a*Hypergeometric2F1[-5/2, 2, -3/2, (1 + Sin[c + d*x])/2])/(d*(a + a*Sin[c + d*x])^(5/2))

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fricas [A]  time = 0.81, size = 187, normalized size = 1.25 \[ \frac {105 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (175 \, \cos \left (d x + c\right )^{2} + 21 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 36\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{960 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(105*sqrt(2)*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c) - 2*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*(175*cos(d*x + c)^2 + 21*(5*
cos(d*x + c)^2 - 4)*sin(d*x + c) - 36)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^
2*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(cos((d*t_nostep+c)/2-pi/4))]Discontinuitie
s at zeroes of cos((d*t_nostep+c)/2-pi/4) were not checkedWarning, integration of abs or sign assumes constant
sign by intervals (correct if the argument is real):Check [abs(t_nostep+1)]Evaluation time: 0.56Not invertibl
e Error: Bad Argument Value

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maple [A]  time = 0.27, size = 124, normalized size = 0.83 \[ \frac {2 a^{3} \left (-\frac {3}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {1}{12 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{20 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}}{2 a \sin \left (d x +c \right )-2 a}-\frac {7 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 \sqrt {a}}}{16 a^{4}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*(-3/16/a^4/(a+a*sin(d*x+c))^(1/2)-1/12/a^3/(a+a*sin(d*x+c))^(3/2)-1/20/a^2/(a+a*sin(d*x+c))^(5/2)-1/16/a
^4*(1/2*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-7/4*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 = 0.75, size = 146, normalized size = 0.97 \[ -\frac {\frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} - 140 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a - 56 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 48 \, a^{3}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}}{960 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(105*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) +
a)))/sqrt(a) + 4*(105*(a*sin(d*x + c) + a)^3 - 140*(a*sin(d*x + c) + a)^2*a - 56*(a*sin(d*x + c) + a)*a^2 - 4
8*a^3)/((a*sin(d*x + c) + a)^(7/2) - 2*(a*sin(d*x + c) + a)^(5/2)*a))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + a*sin(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**3/(a*(sin(c + d*x) + 1))**(3/2), x)

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