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]

-((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]])

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Rubi [A]  time = 0.41596, antiderivative size = 231, normalized size of antiderivative = 1., 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 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 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 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)}{(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*}

Mathematica [C]  time = 0.855788, size = 245, normalized size = 1.06 \[ \frac{-\left (3 a^2 b+3 a^3+7 a b^2+7 b^3\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )+\left (-3 a^2 b+3 a^3+7 a b^2-7 b^3\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )+15 a (a+b) (a+b \sin (c+d x)) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a-b}\right )-3 (a-b) \left (5 a (a+b \sin (c+d x)) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{a+b \sin (c+d x)}{a+b}\right )-2 (a+b) \sec ^2(c+d x) (a \sin (c+d x)-b)\right )}{12 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^{3/2}} \]

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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Maple [A]  time = 0.784, size = 283, normalized size = 1.2 \begin{align*} -{\frac{b}{4\,d \left ( a-b \right ) ^{3} \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 ) ^{3}}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{\frac{7\,b}{4\,d \left ( a-b \right ) ^{3}}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-{\frac{2\,{b}^{3}}{3\,d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-8\,{\frac{a{b}^{3}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}\sqrt{a+b\sin \left ( dx+c \right ) }}}-{\frac{b}{4\,d \left ( a+b \right ) ^{3} \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{7}{2}}}}+{\frac{7\,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{7}{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))^(5/2),x)

[Out]

-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))-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)^(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., 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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \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))^(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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.15749, size = 512, normalized size = 2.22 \begin{align*} \frac{1}{12} \, b^{3}{\left (\frac{3 \,{\left (2 \, a - 7 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a + b}}\right )}{{\left (a^{3} b^{3} d - 3 \, a^{2} b^{4} d + 3 \, a b^{5} d - b^{6} d\right )} \sqrt{-a + b}} - \frac{3 \,{\left (2 \, a + 7 \, b\right )} \arctan \left (\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{\sqrt{-a - b}}\right )}{{\left (a^{3} b^{3} d + 3 \, a^{2} b^{4} d + 3 \, a b^{5} d + b^{6} d\right )} \sqrt{-a - b}} - \frac{6 \,{\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a^{3} - \sqrt{b \sin \left (d x + c\right ) + a} a^{4} + 3 \,{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} a b^{2} - 6 \, \sqrt{b \sin \left (d x + c\right ) + a} a^{2} b^{2} - \sqrt{b \sin \left (d x + c\right ) + a} b^{4}\right )}}{{\left (a^{6} b^{2} d - 3 \, a^{4} b^{4} d + 3 \, a^{2} b^{6} d - b^{8} d\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 )}} - \frac{8 \,{\left (12 \,{\left (b \sin \left (d x + c\right ) + a\right )} a + a^{2} - b^{2}\right )}}{{\left (a^{6} d - 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d - b^{6} d\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\right )} \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))^(5/2),x, algorithm="giac")

[Out]

1/12*b^3*(3*(2*a - 7*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a^3*b^3*d - 3*a^2*b^4*d + 3*a*b^5*d -
b^6*d)*sqrt(-a + b)) - 3*(2*a + 7*b)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a - b))/((a^3*b^3*d + 3*a^2*b^4*d +
3*a*b^5*d + b^6*d)*sqrt(-a - b)) - 6*((b*sin(d*x + c) + a)^(3/2)*a^3 - sqrt(b*sin(d*x + c) + a)*a^4 + 3*(b*si
n(d*x + c) + a)^(3/2)*a*b^2 - 6*sqrt(b*sin(d*x + c) + a)*a^2*b^2 - sqrt(b*sin(d*x + c) + a)*b^4)/((a^6*b^2*d -
3*a^4*b^4*d + 3*a^2*b^6*d - b^8*d)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2 - b^2)) - 8*(12*(
b*sin(d*x + c) + a)*a + a^2 - b^2)/((a^6*d - 3*a^4*b^2*d + 3*a^2*b^4*d - b^6*d)*(b*sin(d*x + c) + a)^(3/2)))