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]

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

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Rubi [A]  time = 0.33, antiderivative size = 186, 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 = {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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 1.20, 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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fricas [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+b*sin(d*x+c))^(3/2),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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

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maple [A]  time = 0.89, size = 250, normalized size = 1.34 \[ -\frac {2 b^{3}}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 d \left (a -b \right )^{2} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 d \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {5 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 d \left (a -b \right )^{2} \sqrt {-a +b}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 d \left (a +b \right )^{2} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 d \left (a +b \right )^{\frac {5}{2}}}+\frac {5 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 d \left (a +b \right )^{\frac {5}{2}}} \]

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]

-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)^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*si
n(d*x+c))^(1/2)/(-a+b)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is 4*a-4*b positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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