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*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]])

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Rubi [A]  time = 0.202478, antiderivative size = 150, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [C]  time = 0.0678499, 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]

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

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Maple [A]  time = 0.166, size = 124, normalized size = 0.8 \begin{align*} 2\,{\frac{{a}^{3}}{d} \left ( -1/16\,{\frac{1}{{a}^{4}} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a\sin \left ( dx+c \right ) -a}}-7/4\,{\frac{\sqrt{2}}{\sqrt{a}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) }-3/16\,{\frac{1}{{a}^{4}\sqrt{a+a\sin \left ( dx+c \right ) }}}-1/12\,{\frac{1}{{a}^{3} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{3/2}}}-1/20\,{\frac{1}{{a}^{2} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{5/2}}} \right ) } \end{align*}

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*(-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)))-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))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.50465, size = 502, normalized size = 3.35 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [A]  time = 1.14058, size = 178, normalized size = 1.19 \begin{align*} -\frac{1}{480} \, a^{3}{\left (\frac{105 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{4} d} + \frac{30 \, \sqrt{a \sin \left (d x + c\right ) + a}}{{\left (a \sin \left (d x + c\right ) - a\right )} a^{4} d} + \frac{4 \,{\left (45 \,{\left (a \sin \left (d x + c\right ) + a\right )}^{2} + 20 \,{\left (a \sin \left (d x + c\right ) + a\right )} a + 12 \, a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} a^{4} d}\right )} \end{align*}

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]

-1/480*a^3*(105*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*sin(d*x + c) + a)/sqrt(-a))/(sqrt(-a)*a^4*d) + 30*sqrt(a*sin
(d*x + c) + a)/((a*sin(d*x + c) - a)*a^4*d) + 4*(45*(a*sin(d*x + c) + a)^2 + 20*(a*sin(d*x + c) + a)*a + 12*a^
2)/((a*sin(d*x + c) + a)^(5/2)*a^4*d))