3.510 \(\int \frac{\sec ^3(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\sec ^2(c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{2 d \left (a^2-b^2\right )}-\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{3/2}}+\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.305468, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2668, 741, 827, 1166, 206} \[ -\frac{\sec ^2(c+d x) (b-a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{2 d \left (a^2-b^2\right )}-\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{3/2}}+\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.493575, size = 176, normalized size = 1.22 \[ \frac{\sqrt{a+b} \left (2 a^2-a b-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )-\sqrt{a-b} \left (\left (2 a^2+a b-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )+2 \sqrt{a+b} \sec ^2(c+d x) (a \sin (c+d x)-b) \sqrt{a+b \sin (c+d x)}\right )}{4 d \sqrt{a-b} \sqrt{a+b} \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b]*(2*a^2 - a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] - Sqrt[a - b]*((2*a^2 + a*b -
3*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 2*Sqrt[a + b]*Sec[c + d*x]^2*(-b + a*Sin[c + d*x])*Sqr
t[a + b*Sin[c + d*x]]))/(4*Sqrt[a - b]*Sqrt[a + b]*(-a^2 + b^2)*d)

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Maple [A]  time = 0.487, size = 218, normalized size = 1.5 \begin{align*} -{\frac{b}{4\,d \left ( a-b \right ) \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 ) }\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{\frac{3\,b}{4\,d \left ( a-b \right ) }\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{\frac{b}{4\,d \left ( a+b \right ) \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{3}{2}}}}+{\frac{3\,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{3}{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))^(1/2),x)

[Out]

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

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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))^(1/2),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))^(1/2),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 )}}{\sqrt{a + b \sin{\left (c + d x \right )}}}\, 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))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.10636, size = 286, normalized size = 1.99 \begin{align*} \frac{b^{3}{\left (\frac{{\left (2 \, a - 3 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a + b}}\right )}{{\left (a b^{3} - b^{4}\right )} \sqrt{-a + b}} - \frac{{\left (2 \, a + 3 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a - b}}\right )}{{\left (a b^{3} + b^{4}\right )} \sqrt{-a - b}} - \frac{2 \,{\left ({\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 )}}{{\left (a^{2} b^{2} - b^{4}\right )}{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{2} - 2 \,{\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}}\right )}}{4 \, d} \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))^(1/2),x, algorithm="giac")

[Out]

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