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

Optimal. Leaf size=186 \[ -\frac{b \left (a^2+5 b^2\right )}{2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{(2 a-5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{5/2}}+\frac{(2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{5/2}} \]

[Out]

-((2*a - 5*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*d) + ((2*a + 5*b)*ArcTanh[Sqrt[a
+ b*Sin[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*d) - (b*(a^2 + 5*b^2))/(2*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c +
d*x]]) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.3343, antiderivative size = 186, 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 = {2668, 741, 829, 827, 1166, 206} \[ -\frac{b \left (a^2+5 b^2\right )}{2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{(2 a-5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{5/2}}+\frac{(2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a - 5*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*d) + ((2*a + 5*b)*ArcTanh[Sqrt[a
+ b*Sin[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*d) - (b*(a^2 + 5*b^2))/(2*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c +
d*x]]) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]])

Rule 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 829

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((e*f - d*g)*(d
+ e*x)^(m + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d*f + a*
e*g - c*(e*f - d*g)*x, x])/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&
FractionQ[m] && LtQ[m, -1]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [C]  time = 1.1061, size = 221, normalized size = 1.19 \[ \frac{\frac{\left (a^2+5 b^2\right ) \left ((a+b) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )\right )}{(a-b) (a+b) \sqrt{a+b \sin (c+d x)}}+\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{\sqrt{a-b}}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b}}+\frac{2 \sec ^2(c+d x) (b-a \sin (c+d x))}{\sqrt{a+b \sin (c+d x)}}}{4 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] - (3*a*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[
a + b]])/Sqrt[a + b] + ((a^2 + 5*b^2)*((a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)] +
(-a + b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)]))/((a - b)*(a + b)*Sqrt[a + b*Sin[c +
d*x]]) + (2*Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/Sqrt[a + b*Sin[c + d*x]])/(4*(-a^2 + b^2)*d)

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Maple [A]  time = 0.545, size = 250, normalized size = 1.3 \begin{align*} -{\frac{b}{4\,d \left ( a-b \right ) ^{2} \left ( b\sin \left ( dx+c \right ) +b \right ) }\sqrt{a+b\sin \left ( dx+c \right ) }}+{\frac{a}{2\,d \left ( a-b \right ) ^{2}}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{\frac{5\,b}{4\,d \left ( a-b \right ) ^{2}}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-2\,{\frac{{b}^{3}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}\sqrt{a+b\sin \left ( dx+c \right ) }}}-{\frac{b}{4\,d \left ( a+b \right ) ^{2} \left ( b\sin \left ( dx+c \right ) -b \right ) }\sqrt{a+b\sin \left ( dx+c \right ) }}+{\frac{a}{2\,d}{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}+{\frac{5\,b}{4\,d}{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*b/(a-b)^2*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2/d/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1
/2)/(-a+b)^(1/2))*a-5/4/d*b/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))-2/d*b^3/(a+b)^2/(
a-b)^2/(a+b*sin(d*x+c))^(1/2)-1/4/d*b/(a+b)^2*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)+1/2/d/(a+b)^(5/2)*arctan
h((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+5/4/d*b/(a+b)^(5/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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 + b \sin{\left (c + d x \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+b*sin(d*x+c))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.11553, size = 401, normalized size = 2.16 \begin{align*} \frac{1}{4} \, b^{3}{\left (\frac{{\left (2 \, a - 5 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a + b}}\right )}{{\left (a^{2} b^{3} d - 2 \, a b^{4} d + b^{5} d\right )} \sqrt{-a + b}} - \frac{{\left (2 \, a + 5 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a - b}}\right )}{{\left (a^{2} b^{3} d + 2 \, a b^{4} d + b^{5} d\right )} \sqrt{-a - b}} - \frac{2 \,{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} a^{2} -{\left (b \sin \left (d x + c\right ) + a\right )} a^{3} + 5 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{2} - 11 \,{\left (b \sin \left (d x + c\right ) + a\right )} a b^{2} + 4 \, a^{2} b^{2} - 4 \, b^{4}\right )}}{{\left (a^{4} b^{2} d - 2 \, a^{2} b^{4} d + b^{6} d\right )}{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} - 2 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a + \sqrt{b \sin \left (d x + c\right ) + a} a^{2} - \sqrt{b \sin \left (d x + c\right ) + a} b^{2}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^3*((2*a - 5*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a^2*b^3*d - 2*a*b^4*d + b^5*d)*sqrt(-a +
b)) - (2*a + 5*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a - b))/((a^2*b^3*d + 2*a*b^4*d + b^5*d)*sqrt(-a - b))
- 2*((b*sin(d*x + c) + a)^2*a^2 - (b*sin(d*x + c) + a)*a^3 + 5*(b*sin(d*x + c) + a)^2*b^2 - 11*(b*sin(d*x + c
) + a)*a*b^2 + 4*a^2*b^2 - 4*b^4)/((a^4*b^2*d - 2*a^2*b^4*d + b^6*d)*((b*sin(d*x + c) + a)^(5/2) - 2*(b*sin(d*
x + c) + a)^(3/2)*a + sqrt(b*sin(d*x + c) + a)*a^2 - sqrt(b*sin(d*x + c) + a)*b^2)))