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

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

[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 = 0.42, 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 = {2668, 741, 829, 827, 1166, 206} \[ -\frac {a b \left (a^2+19 b^2\right )}{2 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}-\frac {b \left (3 a^2+7 b^2\right )}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^{3/2}}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac {(2 a-7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{7/2}}+\frac {(2 a+7 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a - 7*b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(4*(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 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))^{5/2}} \, dx &=\frac {b^3 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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) \operatorname {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) \operatorname {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]  time = 0.88, size = 245, normalized size = 1.06 \[ \frac {-\left (\left (3 a^3+3 a^2 b+7 a b^2+7 b^3\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )\right )+\left (3 a^3-3 a^2 b+7 a b^2-7 b^3\right ) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )+15 a (a+b) (a+b \sin (c+d x)) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a-b}\right )-3 (a-b) \left (5 a (a+b \sin (c+d x)) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a+b \sin (c+d x)}{a+b}\right )-2 (a+b) \sec ^2(c+d x) (a \sin (c+d x)-b)\right )}{12 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^{3/2}} \]

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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fricas [F]  time = 1.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{3}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.93, size = 283, normalized size = 1.23 \[ -\frac {2 b^{3}}{3 d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {8 b^{3} a}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \sqrt {a +b \sin \left (d x +c \right )}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 d \left (a -b \right )^{3} \left (b \sin \left (d x +c \right )+b \right )}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{2 d \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {7 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{4 d \left (a -b \right )^{3} \sqrt {-a +b}}-\frac {b \sqrt {a +b \sin \left (d x +c \right )}}{4 d \left (a +b \right )^{3} \left (b \sin \left (d x +c \right )-b \right )}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{2 d \left (a +b \right )^{\frac {7}{2}}}+\frac {7 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{4 d \left (a +b \right )^{\frac {7}{2}}} \]

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)

[Out]

-2/3/d*b^3/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))^(3/2)-8/d*b^3*a/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))^(1/2)-1/4/d*b/(a-
b)^3*(a+b*sin(d*x+c))^(1/2)/(b*sin(d*x+c)+b)+1/2/d/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(
1/2))*a-7/4/d*b/(a-b)^3/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))-1/4/d*b/(a+b)^3*(a+b*sin(d*x+
c))^(1/2)/(b*sin(d*x+c)-b)+1/2/d/(a+b)^(7/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+7/4/d*b/(a+b)^(7/2)
*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))^(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 details)Is 4*a-4*b positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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