3.6.31 \(\int \frac {\sec ^3(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [531]

Optimal. Leaf size=231 \[ -\frac {(2 a-7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{7/2} d}+\frac {(2 a+7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{7/2} d}-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

-1/4*(2*a-7*b)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(7/2)/d+1/4*(2*a+7*b)*arctanh((a+b*sin(d*x+c)
)^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/d-1/6*b*(3*a^2+7*b^2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(3/2)-1/2*sec(d*x+c)^2*(
b-a*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))^(3/2)-1/2*a*b*(a^2+19*b^2)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2747, 755, 843, 841, 1180, 212} \begin {gather*} -\frac {a b \left (a^2+19 b^2\right )}{2 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}-\frac {b \left (3 a^2+7 b^2\right )}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac {(2 a-7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{7/2}}+\frac {(2 a+7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

-1/4*((2*a - 7*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/((a - b)^(7/2)*d) + ((2*a + 7*b)*ArcTanh[Sqrt
[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(7/2)*d) - (b*(3*a^2 + 7*b^2))/(6*(a^2 - b^2)^2*d*(a + b*Sin[c +
d*x])^(3/2)) - (Sec[c + d*x]^2*(b - a*Sin[c + d*x]))/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(3/2)) - (a*b*(a^2
+ 19*b^2))/(2*(a^2 - b^2)^3*d*Sqrt[a + b*Sin[c + d*x]])

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 843

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 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)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac {b^3 \text {Subst}\left (\int \frac {1}{(a+x)^{5/2} \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 (a+b \sin (c+d x))^{3/2}}+\frac {b \text {Subst}\left (\int \frac {\frac {1}{2} \left (2 a^2-7 b^2\right )+\frac {5 a x}{2}}{(a+x)^{5/2} \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 (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {b \text {Subst}\left (\int \frac {-a \left (a^2-6 b^2\right )-\frac {1}{2} \left (3 a^2+7 b^2\right ) x}{(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 )^2 d}\\ &=-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\frac {b \text {Subst}\left (\int \frac {\frac {1}{2} \left (2 a^4-15 a^2 b^2-7 b^4\right )+\frac {1}{2} a \left (a^2+19 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 )^3 d}\\ &=-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}+\frac {b \text {Subst}\left (\int \frac {-\frac {1}{2} a^2 \left (a^2+19 b^2\right )+\frac {1}{2} \left (2 a^4-15 a^2 b^2-7 b^4\right )+\frac {1}{2} a \left (a^2+19 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 )^3 d}\\ &=-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {(2 a-7 b) \text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 (a-b)^3 d}+\frac {(2 a+7 b) \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 (a+b)^3 d}\\ &=-\frac {(2 a-7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{7/2} d}+\frac {(2 a+7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{7/2} d}-\frac {b \left (3 a^2+7 b^2\right )}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {a b \left (a^2+19 b^2\right )}{2 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.87, size = 245, normalized size = 1.06 \begin {gather*} \frac {-\left (\left (3 a^3+3 a^2 b+7 a b^2+7 b^3\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )\right )+\left (3 a^3-3 a^2 b+7 a b^2-7 b^3\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )+15 a (a+b) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))-3 (a-b) \left (-2 (a+b) \sec ^2(c+d x) (-b+a \sin (c+d x))+5 a \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))\right )}{12 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-((3*a^3 + 3*a^2*b + 7*a*b^2 + 7*b^3)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a - b)]) + (3*a^
3 - 3*a^2*b + 7*a*b^2 - 7*b^3)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[c + d*x])/(a + b)] + 15*a*(a + b)*H
ypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a - b)]*(a + b*Sin[c + d*x]) - 3*(a - b)*(-2*(a + b)*Sec[
c + d*x]^2*(-b + a*Sin[c + d*x]) + 5*a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[c + d*x])/(a + b)]*(a + b*Si
n[c + d*x])))/(12*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [A]
time = 2.73, size = 263, normalized size = 1.14

method	result	size
default	\(\frac {-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 \left (a +b \right )^{\frac {7}{2}}}+\frac {7 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 \left (a +b \right )^{\frac {7}{2}}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {7 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {2 b^{3}}{3 \left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {8 b^{3} a}{\left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a +b \sin \left (d x +c \right )}}}{d}\)	\(263\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-1/4*b/(a+b)^3*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)-b)+1/2/(a+b)^(7/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(
1/2))*a+7/4*b/(a+b)^(7/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))-1/4*b/(a-b)^3*(a+b*sin(d*x+c))^(1/2)/(b*
sin(d*x+c)+b)+1/2/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a-7/4*b/(a-b)^3/(-a+b)^(1/2
)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))-2/3*b^3/(a-b)^2/(a+b)^2/(a+b*sin(d*x+c))^(3/2)-8*b^3*a/(a-b)^3/(
a+b)^3/(a+b*sin(d*x+c))^(1/2))/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))^(5/2),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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^3/(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^
2 - 3*a^2*b - b^3)*sin(d*x + c)), x)

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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 )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, 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))**(5/2),x)

[Out]

Integral(sec(c + d*x)**3/(a + b*sin(c + d*x))**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))^(5/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,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))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + b*sin(c + d*x))^(5/2)), x)

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