3.5.88 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) [488]

Optimal. Leaf size=188 \[ -\frac {3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 \sqrt {a-b} d}+\frac {3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 \sqrt {a+b} d}-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d} \]

[Out]

-3/32*(4*a^2-2*a*b-b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d/(a-b)^(1/2)+3/32*(4*a^2+2*a*b-b^2)*arcta
nh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/d/(a+b)^(1/2)-1/16*sec(d*x+c)^2*(b-6*a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/
2)/d+1/4*sec(d*x+c)^4*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.20, antiderivative size = 188, 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 = {2747, 753, 837, 841, 1180, 212} \begin {gather*} -\frac {3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d \sqrt {a-b}}+\frac {3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d \sqrt {a+b}}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b) \sqrt {a+b \sin (c+d x)}}{4 d}-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*a^2 - 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*Sqrt[a - b]*d) + (3*(4*a^2 + 2*a*
b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*Sqrt[a + b]*d) - (Sec[c + d*x]^2*(b - 6*a*Sin[c +
d*x])*Sqrt[a + b*Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^4*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(4*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 753

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*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 \sec ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac {b^5 \text {Subst}\left (\int \frac {(a+x)^{3/2}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d}-\frac {b^3 \text {Subst}\left (\int \frac {\frac {1}{2} \left (-6 a^2+b^2\right )-\frac {5 a x}{2}}{\sqrt {a+x} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d}+\frac {b \text {Subst}\left (\int \frac {\frac {3}{4} \left (4 a^4-5 a^2 b^2+b^4\right )+\frac {3}{2} a \left (a^2-b^2\right ) x}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d}+\frac {b \text {Subst}\left (\int \frac {-\frac {3}{2} a^2 \left (a^2-b^2\right )+\frac {3}{4} \left (4 a^4-5 a^2 b^2+b^4\right )+\frac {3}{2} a \left (a^2-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 )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d}-\frac {\left (3 \left (4 a^2-2 a b-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 d}+\frac {\left (3 \left (4 a^2+2 a b-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 d}\\ &=-\frac {3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 \sqrt {a-b} d}+\frac {3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 \sqrt {a+b} d}-\frac {\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{16 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d}\\ \end {align*}

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Mathematica [A]
time = 2.49, size = 297, normalized size = 1.58 \begin {gather*} -\frac {3 \sqrt {a-b} (a+b)^2 \left (4 a^3-6 a^2 b+a b^2+b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )-3 (a-b)^2 \sqrt {a+b} \left (4 a^3+6 a^2 b+a b^2-b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+8 \left (-a^2+b^2\right ) \sec ^4(c+d x) (-b+a \sin (c+d x)) (a+b \sin (c+d x))^{5/2}+2 \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \left (5 a^2 b-3 b^3+\left (-6 a^3+4 a b^2\right ) \sin (c+d x)\right )-2 b \sqrt {a+b \sin (c+d x)} \left (12 a^4-13 a^2 b^2+3 b^4+\left (6 a^3 b-4 a b^3\right ) \sin (c+d x)\right )}{32 \left (a^2-b^2\right )^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(3*Sqrt[a - b]*(a + b)^2*(4*a^3 - 6*a^2*b + a*b^2 + b^3)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] -
3*(a - b)^2*Sqrt[a + b]*(4*a^3 + 6*a^2*b + a*b^2 - b^3)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 8*(-a
^2 + b^2)*Sec[c + d*x]^4*(-b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(5/2) + 2*Sec[c + d*x]^2*(a + b*Sin[c + d*
x])^(5/2)*(5*a^2*b - 3*b^3 + (-6*a^3 + 4*a*b^2)*Sin[c + d*x]) - 2*b*Sqrt[a + b*Sin[c + d*x]]*(12*a^4 - 13*a^2*
b^2 + 3*b^4 + (6*a^3*b - 4*a*b^3)*Sin[c + d*x]))/((a^2 - b^2)^2*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(408\) vs. \(2(164)=328\).
time = 2.05, size = 409, normalized size = 2.18

method	result	size
default	\(\frac {4 \sqrt {a +b}\, \sqrt {-a +b}\, \sqrt {a +b \sin \left (d x +c \right )}\, b \left (b \left (\cos ^{2}\left (d x +c \right )\right )+8 a \sin \left (d x +c \right )-b \right )+3 b \left (4 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2} \sqrt {a +b}-2 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a \sqrt {a +b}-b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) \sqrt {a +b}+4 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2} \sqrt {-a +b}+2 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a \sqrt {-a +b}-b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) \sqrt {-a +b}\right ) \left (\cos ^{4}\left (d x +c \right )\right )+6 \sqrt {a +b}\, \sqrt {-a +b}\, \sqrt {a +b \sin \left (d x +c \right )}\, b \left (2 a \sin \left (d x +c \right )-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )-24 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a \sqrt {a +b}\, \sqrt {-a +b}+24 \sqrt {a +b \sin \left (d x +c \right )}\, a^{2} \sqrt {a +b}\, \sqrt {-a +b}+12 \sqrt {a +b \sin \left (d x +c \right )}\, b^{2} \sqrt {a +b}\, \sqrt {-a +b}}{32 \sqrt {a +b}\, \sqrt {-a +b}\, b \cos \left (d x +c \right )^{4} d}\)	\(409\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*(a+b)^(1/2)*(-a+b)^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(b*cos(d*x+c)^2+8*a*sin(d*x+c)-b)+3*b*(4*arctan((a+b
*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2*(a+b)^(1/2)-2*b*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a*(a+b)^(1/2)
-b^2*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*(a+b)^(1/2)+4*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2
*(-a+b)^(1/2)+2*b*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*(-a+b)^(1/2)-b^2*arctanh((a+b*sin(d*x+c))^(1/2
)/(a+b)^(1/2))*(-a+b)^(1/2))*cos(d*x+c)^4+6*(a+b)^(1/2)*(-a+b)^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(2*a*sin(d*x+c)-
b)*cos(d*x+c)^2-24*(a+b*sin(d*x+c))^(3/2)*a*(a+b)^(1/2)*(-a+b)^(1/2)+24*(a+b*sin(d*x+c))^(1/2)*a^2*(a+b)^(1/2)
*(-a+b)^(1/2)+12*(a+b*sin(d*x+c))^(1/2)*b^2*(a+b)^(1/2)*(-a+b)^(1/2))/(a+b)^(1/2)/(-a+b)^(1/2)/b/cos(d*x+c)^4/
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)^5*(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 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)^5*(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^5*sin(d*x + c) + a*sec(d*x + c)^5)*sqrt(b*sin(d*x + c) + a), x)

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Sympy [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)**5*(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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