3.6.11 \(\int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [511]

Optimal. Leaf size=230 \[ -\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{5/2} d}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{5/2} d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d} \]

[Out]

-3/32*(4*a^2-10*a*b+7*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+3/32*(4*a^2+10*a*b+7*b^2)
*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^
(1/2)/(a^2-b^2)/d-1/16*sec(d*x+c)^2*(b*(a^2-7*b^2)-6*a*(a^2-2*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2
)^2/d

________________________________________________________________________________________

Rubi [A]
time = 0.25, antiderivative size = 230, 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, 755, 837, 841, 1180, 212} \begin {gather*} -\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d (a-b)^{5/2}}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d (a+b)^{5/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-3*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(5/2)*d) + (3*(4*a^2 +
10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(5/2)*d) - (Sec[c + d*x]^4*(b - a*
Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*(b*(a^2 -
7*b^2) - 6*a*(a^2 - 2*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)^2*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 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 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 \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {b^5 \text {Subst}\left (\int \frac {1}{\sqrt {a+x} \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 \left (a^2-b^2\right ) d}+\frac {b^3 \text {Subst}\left (\int \frac {\frac {1}{2} \left (6 a^2-7 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 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {b \text {Subst}\left (\int \frac {-\frac {3}{4} \left (4 a^4-9 a^2 b^2+7 b^4\right )-\frac {3}{2} a \left (a^2-2 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 )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {b \text {Subst}\left (\int \frac {\frac {3}{2} a^2 \left (a^2-2 b^2\right )-\frac {3}{4} \left (4 a^4-9 a^2 b^2+7 b^4\right )-\frac {3}{2} a \left (a^2-2 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 )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {\left (3 \left (4 a^2-10 a b+7 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 (a-b)^2 d}+\frac {\left (3 \left (4 a^2+10 a b+7 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 (a+b)^2 d}\\ &=-\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{5/2} d}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{5/2} d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.90, size = 244, normalized size = 1.06 \begin {gather*} \frac {-3 (a+b)^{5/2} \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+\sqrt {a-b} \left (3 (a-b)^2 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \left (-9 a^2 b+15 b^3+\left (-a^2 b+7 b^3\right ) \cos (2 (c+d x))+a \left (11 a^2-14 b^2\right ) \sin (c+d x)+3 \left (a^3-2 a b^2\right ) \sin (3 (c+d x))\right )\right )}{32 \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-3*(a + b)^(5/2)*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] + Sqrt[a - b]*(3*(a -
b)^2*(4*a^2 + 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + Sqrt[a + b]*Sec[c + d*x]^4*Sqrt
[a + b*Sin[c + d*x]]*(-9*a^2*b + 15*b^3 + (-(a^2*b) + 7*b^3)*Cos[2*(c + d*x)] + a*(11*a^2 - 14*b^2)*Sin[c + d*
x] + 3*(a^3 - 2*a*b^2)*Sin[3*(c + d*x)])))/(32*Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)^2*d)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(579\) vs. \(2(206)=412\).
time = 3.43, size = 580, normalized size = 2.52

method	result	size
default	\(\frac {-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {9 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a}{16 \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a +b \right )}+\frac {11 b^{2} \sqrt {a +b \sin \left (d x +c \right )}}{32 \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a +b \right )}+\frac {3 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}+\frac {15 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a b}{16 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}+\frac {21 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) b^{2}}{32 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {9 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a}{16 \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a -b \right )}-\frac {11 b^{2} \sqrt {a +b \sin \left (d x +c \right )}}{32 \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a -b \right )}+\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{8 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}-\frac {15 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a b}{16 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}+\frac {21 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) b^{2}}{32 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}}{d}\)	\(580\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-3/16/(b*sin(d*x+c)-b)^2*b/(a^2+2*a*b+b^2)*(a+b*sin(d*x+c))^(3/2)*a-9/32/(b*sin(d*x+c)-b)^2*b^2/(a^2+2*a*b+b^
2)*(a+b*sin(d*x+c))^(3/2)+3/16/(b*sin(d*x+c)-b)^2*b/(a+b)*(a+b*sin(d*x+c))^(1/2)*a+11/32/(b*sin(d*x+c)-b)^2*b^
2/(a+b)*(a+b*sin(d*x+c))^(1/2)+3/8/(a^2+2*a*b+b^2)/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2
+15/16/(a^2+2*a*b+b^2)/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*b+21/32/(a^2+2*a*b+b^2)/(a+b)
^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*b^2-3/16/(b*sin(d*x+c)+b)^2*b/(a^2-2*a*b+b^2)*(a+b*sin(d*x+
c))^(3/2)*a+9/32/(b*sin(d*x+c)+b)^2*b^2/(a^2-2*a*b+b^2)*(a+b*sin(d*x+c))^(3/2)+3/16/(b*sin(d*x+c)+b)^2*b/(a-b)
*(a+b*sin(d*x+c))^(1/2)*a-11/32/(b*sin(d*x+c)+b)^2*b^2/(a-b)*(a+b*sin(d*x+c))^(1/2)+3/8/(a^2-2*a*b+b^2)/(-a+b)
^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2-15/16/(a^2-2*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+
c))^(1/2)/(-a+b)^(1/2))*a*b+21/32/(a^2-2*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*b^2
)/d

________________________________________________________________________________________

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))^(1/2),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

________________________________________________________________________________________

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))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \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))**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (208) = 416\).
time = 4.70, size = 419, normalized size = 1.82 \begin {gather*} \frac {b^{5} {\left (\frac {3 \, {\left (4 \, a^{2} - 10 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a + b}}\right )}{{\left (a^{2} b^{5} - 2 \, a b^{6} + b^{7}\right )} \sqrt {-a + b}} - \frac {3 \, {\left (4 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{\sqrt {-a - b}}\right )}{{\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} \sqrt {-a - b}} - \frac {2 \, {\left (6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + 18 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} - 6 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{6} - 12 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a b^{2} + 35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} b^{2} - 44 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} b^{2} + 21 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{4} b^{2} + 7 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} b^{4} + 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a b^{4} - 4 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{2} b^{4} - 11 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{6}\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\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 )}^{2}}\right )}}{32 \, d} \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))^(1/2),x, algorithm="giac")

[Out]

1/32*b^5*(3*(4*a^2 - 10*a*b + 7*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a + b))/((a^2*b^5 - 2*a*b^6 + b^7)*
sqrt(-a + b)) - 3*(4*a^2 + 10*a*b + 7*b^2)*arctan(sqrt(b*sin(d*x + c) + a)/sqrt(-a - b))/((a^2*b^5 + 2*a*b^6 +
b^7)*sqrt(-a - b)) - 2*(6*(b*sin(d*x + c) + a)^(7/2)*a^3 - 18*(b*sin(d*x + c) + a)^(5/2)*a^4 + 18*(b*sin(d*x
+ c) + a)^(3/2)*a^5 - 6*sqrt(b*sin(d*x + c) + a)*a^6 - 12*(b*sin(d*x + c) + a)^(7/2)*a*b^2 + 35*(b*sin(d*x + c
) + a)^(5/2)*a^2*b^2 - 44*(b*sin(d*x + c) + a)^(3/2)*a^3*b^2 + 21*sqrt(b*sin(d*x + c) + a)*a^4*b^2 + 7*(b*sin(
d*x + c) + a)^(5/2)*b^4 + 2*(b*sin(d*x + c) + a)^(3/2)*a*b^4 - 4*sqrt(b*sin(d*x + c) + a)*a^2*b^4 - 11*sqrt(b*
sin(d*x + c) + a)*b^6)/((a^4*b^4 - 2*a^2*b^6 + b^8)*((b*sin(d*x + c) + a)^2 - 2*(b*sin(d*x + c) + a)*a + a^2 -
b^2)^2))/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________