3.6.10 \(\int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [510]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 144, normalized size of antiderivative = 1.00, 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 = {2747, 755, 841, 1180, 212} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/((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 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 755

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 841

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 1180

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 2747

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]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {b^3 \text {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 \text {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 \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.54, size = 176, normalized size = 1.22 \begin {gather*} \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) (-b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}\right )}{4 \sqrt {a-b} \sqrt {a+b} \left (-a^2+b^2\right ) d} \end {gather*}

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 = 2.31, size = 204, normalized size = 1.42

method	result	size
default	\(\frac {-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a +b \right ) \left (b \sin \left (d x +c \right )-b \right )}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 \left (a +b \right )^{\frac {3}{2}}}+\frac {3 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) b}{4 \left (a +b \right )^{\frac {3}{2}}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a -b \right ) \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 \left (a -b \right ) \sqrt {-a +b}}-\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) b}{4 \left (a -b \right ) \sqrt {-a +b}}}{d}\)	\(204\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more detail

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >>  Error detected within library code:   integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \end {gather*}

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 = 2.21, size = 212, normalized size = 1.47 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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