Optimal. Leaf size=296 \[ -\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^2 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d} \]
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Rubi [A]
time = 0.76, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4187, 4191,
3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac {2 (b B-a C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b^2 d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac {2 C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3856
Rule 3872
Rule 3934
Rule 4187
Rule 4191
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {2 \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a C}{2}+\frac {1}{2} b (5 A+3 C) \sec (c+d x)+\frac {5}{2} (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{5 b}\\ &=\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\sqrt {\sec (c+d x)} \left (\frac {5}{4} a (b B-a C)+\frac {1}{4} b (5 b B+4 a C) \sec (c+d x)+\frac {3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{15 b^2}\\ &=\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {8 \int \frac {-\frac {3}{8} a \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\frac {1}{8} b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \sec (c+d x)+\frac {5}{8} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 b^3}\\ &=\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {8 \int \frac {-\frac {3}{8} a^2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\left (-\frac {3}{8} a b \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )+\frac {1}{8} a b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^2 b^3}-\frac {\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{b^3}\\ &=\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {(b B-a C) \int \sqrt {\sec (c+d x)} \, dx}{3 b^2}-\frac {\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{5 b^3}-\frac {\left (a \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3}\\ &=-\frac {2 a \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {\left ((b B-a C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^2}-\frac {\left (\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^2 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b B-a C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 C \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}
Verification is not applicable to the result.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(772\) vs.
\(2(346)=692\).
time = 0.27, size = 773, normalized size = 2.61
method | result | size |
default | \(\text {Expression too large to display}\) | \(773\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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